| Term | Description | Symbol |
|---|---|---|
| Optical Centre/Pole | Reference point for all measurements | C/O (lens), P (mirror) |
| Principal Axis | Imaginary line through C/P | — |
| Principal Focus | Point where parallel rays converge/diverge from | F |
| Focal Length | Distance from C/P to F | f |
| Object Distance | Distance of object from C/P | u |
| Image Distance | Distance of image from C/P | v |
Welcome, future physicists and engineers! In this deep dive, we're going to unravel the intricacies of determining the focal length of optical elements – specifically, a convex lens and a convex mirror – using the powerful u-v method. This method is not just about memorizing formulas; it's about understanding the fundamental principles of optics, mastering experimental techniques, and skillfully interpreting data, all crucial for success in JEE.
The 'u-v method' fundamentally refers to measuring the object distance (u) and the corresponding image distance (v) for various positions of the object, and then utilizing the lens or mirror formula to determine the focal length (f). It's a cornerstone experiment in ray optics.
A convex lens, also known as a converging lens, forms real and inverted images for objects placed beyond its focal point. This property allows us to use a screen to locate the image, making the u-v method highly practical.
The foundation of this method is the Thin Lens Formula:
$$frac{1}{f} = frac{1}{v} - frac{1}{u}$$
where:
For accurate application, the Cartesian Sign Convention is paramount for JEE. Let's briefly recap:
For a convex lens forming a real image of a real object, typically:
So, the lens formula for magnitudes becomes: $$frac{1}{f} = frac{1}{|v|} + frac{1}{|u|}$$
However, it is always safer to stick to the original formula with signed values: $$frac{1}{f} = frac{1}{v} - frac{1}{u}$$ and ensure you apply signs consistently.
The experiment is typically performed on an optical bench. The key components are:
Procedure Steps:
While averaging the calculated 'f' values gives an estimate, graphical methods are superior for minimizing random errors and are frequently tested in JEE.
If you plot 'v' on the y-axis and 'u' on the x-axis (taking magnitudes), you'll get a hyperbolic curve.
From the lens formula: $$frac{1}{f} = frac{1}{v} - frac{1}{u}$$
If we consider magnitudes and let u be positive (object distance) and v be positive (real image distance): $$frac{1}{f} = frac{1}{v} + frac{1}{u}$$
Rearranging: $$v = frac{fu}{u-f}$$
Key points on the graph:
From the point (2f, 2f), you can directly read the value of 2f and hence determine 'f'.
This is the most accurate and commonly asked graphical method. Let's rewrite the lens formula keeping the Cartesian sign convention in mind:
$$frac{1}{f} = frac{1}{v} - frac{1}{u}$$
For a real object (u negative) and a real image (v positive) formed by a convex lens:
Let $u_{measured}$ be the magnitude of object distance and $v_{measured}$ be the magnitude of image distance.
So, $u = -u_{measured}$ and $v = +v_{measured}$.
Substituting into the formula: $$frac{1}{f} = frac{1}{v_{measured}} - frac{1}{(-u_{measured})} = frac{1}{v_{measured}} + frac{1}{u_{measured}}$$
If we plot $frac{1}{v_{measured}}$ on the y-axis and $frac{1}{u_{measured}}$ on the x-axis:
$$frac{1}{v_{measured}} = -frac{1}{u_{measured}} + frac{1}{f}$$
This equation is in the form y = mx + c, representing a straight line.
By finding the intercepts, you can directly determine f (f = 1/intercept). Averaging the inverse of both intercepts provides a more accurate value of 'f'.
JEE Advanced Tip: Error Analysis
Understanding how errors propagate is crucial. For the lens formula $frac{1}{f} = frac{1}{v} - frac{1}{u}$ (using signed values), the fractional error in 'f' can be estimated. Taking differentials:
$-frac{df}{f^2} = -frac{dv}{v^2} + frac{du}{u^2}$
Considering maximum possible error (absolute values):
$frac{Delta f}{f^2} = frac{Delta v}{v^2} + frac{Delta u}{u^2}$
This shows that errors in measuring u or v (Δu, Δv) lead to larger errors in 'f' when u or v are small (close to f). The percentage error is minimum when u = v = 2f, making this a preferred region for measurements in practical scenarios.
| Object Position (u) | Image Position (v) | Nature of Image | Size of Image |
|---|---|---|---|
| At Infinity | At F (focus) | Real, Inverted | Highly Diminished (point size) |
| Beyond 2F | Between F and 2F | Real, Inverted | Diminished |
| At 2F | At 2F | Real, Inverted | Same Size |
| Between F and 2F | Beyond 2F | Real, Inverted | Magnified |
| At F | At Infinity | Real, Inverted | Highly Magnified |
| Between F and Optical Centre | Same side as object | Virtual, Erect | Magnified |
The u-v method is primarily concerned with obtaining real images (u > f) to enable screen localization. For virtual images (u < f), indirect methods or observation through the lens are required.
A convex mirror always forms virtual, erect, and diminished images for real objects. This means a screen cannot be used directly to find the image position (v) as with a convex lens. Therefore, a direct 'u-v' measurement for real objects is not feasible. To use the u-v method (i.e., the mirror formula) for a convex mirror, we need to employ an indirect approach, usually involving an auxiliary lens or creating a virtual object for the convex mirror.
The fundamental equation is the Mirror Formula:
$$frac{1}{f} = frac{1}{v} + frac{1}{u}$$
Again, the Cartesian Sign Convention is critical:
Substituting these signs for a real object and its virtual image:
$$frac{1}{-|f|} = frac{1}{|v|} + frac{1}{-|u|} implies frac{1}{-|f|} = frac{1}{|v|} - frac{1}{|u|}$$
This means for magnitudes:
$$frac{1}{|f|} = frac{1}{|u|} - frac{1}{|v|}$$
However, the direct u-v method often relies on forming a real image. For a convex mirror, this requires a virtual object.
This is the primary method to apply the u-v concept to a convex mirror, as it allows for the formation of a real image, which can then be located on a screen.
The core idea is to use an auxiliary convex lens (L) to form a real image (I1) of a real object (O). This image (I1) then acts as a virtual object for the convex mirror (M). By carefully placing the convex mirror, we can make it form a final real image (I2) of this virtual object, which can be located on a screen.
Let's consider the mirror formula: $$frac{1}{f_M} = frac{1}{v_M} + frac{1}{u_M}$$
Using the Cartesian sign convention:
So, the formula becomes (using magnitudes $|u_M|$ and $|v_M|$):
$$frac{1}{f_M} = frac{1}{-|v_M|} + frac{1}{|u_M|}$$
Which can be written as:
$$frac{1}{f_M} = frac{1}{|u_M|} - frac{1}{|v_M|}$$
Plotting $frac{1}{|v_M|}$ on the y-axis against $frac{1}{|u_M|}$ on the x-axis:
Rearranging the formula: $$frac{1}{|v_M|} = frac{1}{|u_M|} - frac{1}{f_M}$$
This is again a straight line of the form y = mx + c:
From the x-intercept, you can directly find $f_M$. From the y-intercept, you can find $f_M = -1/( ext{y-intercept})$. Both should give the same positive value for the focal length of the convex mirror.
CBSE vs. JEE Focus:
By thoroughly understanding these principles, mastering the sign conventions, and practicing graphical analysis, you'll be well-prepared for any u-v method problem involving lenses or mirrors in your JEE journey!
Mastering the 'u-v method' for determining focal lengths often boils down to remembering key formulas, sign conventions, and graphical interpretations. Here are some mnemonics and short-cuts to help you recall these critical points quickly and accurately in exams.
One of the most common errors is confusing the positive/negative sign in the mirror and lens formulas.
| Device | Formula | Sign of Operation |
|---|---|---|
| Mirror | 1/f = 1/v + 1/u | (+) "Adds" |
| Lens | 1/f = 1/v – 1/u | (–) "Subtracts" |
Correctly applying sign conventions is paramount. Remember these fundamental rules:
The sign of the focal length directly tells you if the device is converging or diverging.
| Device Type | Examples | Focal Length (f) |
|---|---|---|
| Converging | Convex Lens, Concave Mirror | Positive (+) |
| Diverging | Concave Lens, Convex Mirror | Negative (-) |
This general rule helps in cross-checking your results for 'v' and magnification (m).
For experimental determination of focal length, plotting 1/v versus 1/u is very common.
For a convex lens, there's a unique point on the u-v graph.
In essence, mastering the concepts behind focal length determination through methods like u-v analysis provides the foundational knowledge necessary for the design, manufacture, and application of almost every optical device in our modern world.
How it works: To get a sharp, focused image on the screen, you often have to adjust the distance of the projector from the screen (changing 'v'). If you move the projector closer or further from the screen, you'll notice the image becomes blurry. To regain focus, you adjust the lens within the projector (which effectively changes the 'u' slightly or shifts the internal elements). The projector lens has a fixed focal length 'f', and for any given 'u', there's only one specific 'v' at which a clear image can be formed, as dictated by the lens/mirror formula. The u–v method systematically explores these pairs to determine 'f'.
How it works: When you "focus" your camera on an object, you are essentially adjusting the distance between the lens and the sensor/film (changing 'v') until a sharp image is formed on the sensor for that particular object distance 'u'. For a lens with a fixed focal length 'f', only specific combinations of 'u' and 'v' will result in a perfectly sharp, in-focus image. The u–v method in experiments is exactly about finding these precise 'u' and 'v' pairs that yield a sharp image and then using them to calculate 'f'.
To effectively understand and perform the experimental determination of focal length using the u–v method for convex lenses and mirrors, a solid grasp of fundamental geometrical optics concepts is essential. These prerequisites form the theoretical backbone for the practical setup and calculations involved.
Mastering these foundational concepts will make learning and performing the u–v method not just easier, but also more meaningful, enabling you to grasp the underlying physics rather than just memorizing steps.
When dealing with the u-v method for determining the focal length of convex lenses and mirrors, students often fall into specific traps during exams. Mastering these nuances can significantly improve accuracy and prevent common errors.
By being mindful of these common traps, you can approach questions related to the u-v method with greater confidence and accuracy in both JEE Main and CBSE board examinations.
While calculating 'f' from individual (u,v) pairs and averaging is possible, graphical methods are preferred for accuracy and to account for experimental errors.
| Aspect | CBSE Board Exams | JEE Main/Advanced |
|---|---|---|
| Emphasis | Procedural accuracy, observation tables, neat graphs, calculation of 'f' and percentage error. | Conceptual understanding, error analysis (least count, propagation), application of formulas, interpretation of graphical results, understanding of limitations. |
| Key Skills | Measurement, plotting, basic calculations. | Critical thinking, problem-solving with experimental data, error estimation, selection of best method. |
JEE Specific Tip: Be prepared for questions involving uncertainty in measurements (e.g., least count of the scale) and how it propagates to the focal length calculation. Understanding the significance of the "2f" point in both graphs and physical setup is vital.
Mastering this experiment not only helps in scoring well but also builds a strong foundation in practical optics!
Mastering the u–v method for determining focal length is crucial for both theoretical numerical problems and interpreting experimental data. A systematic approach minimizes errors and helps in quick, accurate solutions.
| Parameter | Convex Lens | Convex Mirror |
|---|---|---|
| Object Distance (u) | Always negative (real object on left) | Always negative (real object on left) |
| Image Distance (v) | Positive for real image (right) Negative for virtual image (left) | Always positive (virtual image behind mirror) |
| Focal Length (f) | Always positive (converging) | Always positive (diverging, but focal point is virtual) |
JEE Tip: For concave lenses/mirrors, focal length (f) is negative. Ensure you correctly identify the type!
1/v - 1/u = 1/f1/v + 1/u = 1/fm = hi/ho = v/um = hi/ho = -v/u
The u–v method often involves graphical analysis, especially for JEE.
(1/v) = (1/u) + (1/f). This is a straight line of the form y = mx + c, with slope m = 1 and y-intercept c = 1/f. The x-intercept will be -1/f.(1/v) = -(1/u) + (1/f). This is a straight line with slope m = -1 and y-intercept c = 1/f. The x-intercept will also be 1/f.u = v = 2f (for real, inverted image of same size) is a key point on this graph for a lens, or u = v = f for a mirror (virtual image).
Stay meticulous with sign conventions – they are the cornerstone of solving optics problems!
For CBSE board examinations, the experiment to determine the focal length of a convex lens and a concave mirror using the u-v method is a cornerstone of the practical syllabus. Emphasis is placed on understanding the underlying principles, meticulous experimental procedure, accurate data recording, and correct graphical analysis. Mastery of sign conventions is paramount.
CBSE places significant importance on correct graph plotting and interpretation.
Expect questions on:
JEE Main Callout: While CBSE focuses on practical execution and basic understanding, JEE Main might test your conceptual grasp of the lens/mirror formula, sign conventions, graph interpretation, and the ability to solve problems involving object/image distances and focal length, often in multiple-choice format, without direct experimental setup.
This experimental skill forms a fundamental part of Ray Optics and is frequently tested in JEE Main, not just as a direct experimental question, but also integrated into conceptual problems. Mastery of sign conventions and graphical analysis is paramount.
Understanding the correct formula and consistently applying sign conventions is the bedrock of solving problems related to this topic.
$ frac{1}{v} - frac{1}{u} = frac{1}{f} $
For a convex lens, 'f' is positive. Real images form when object is beyond F, and are inverted. Virtual images form when object is between O and F, and are erect.
$ frac{1}{v} + frac{1}{u} = frac{1}{f} $
For a concave mirror, 'f' is negative. Real images form when object is beyond F (and at F), and are inverted. Virtual images form when object is between P and F, and are erect.
JEE often tests the interpretation of graphs derived from the lens/mirror formula. These methods provide alternative ways to determine focal length.
| Aspect | CBSE Board Exams | JEE Main |
|---|---|---|
| Focus | Procedure, observation table, calculations, basic graph plotting. | Conceptual understanding, intricate graphical analysis (e.g., intercepts, slopes), application of sign conventions in complex scenarios, error analysis (indirectly), variations in experimental setup. |
| Problem Types | Direct application of formula, tabulation of u-v values and calculating 'f', simple plotting. | Interpreting graphs to find 'f', questions involving magnification and image properties, scenarios with multiple lenses/mirrors, conditions for real/virtual images. |
Motivation: Mastering the u-v method and its graphical representations not only helps in scoring well in experimental skills but also strengthens your conceptual foundation for advanced optics problems. Practice diverse problems, especially those involving graphs, to build confidence!
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A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
A student plots $1/v$ (Y) vs $1/u$ (X) for a convex lens. The graph intersects the X-axis at $0.04 ext{ cm}^{-1}$.
Wrong Calculation: Student assumes $1/f = 0.04 implies f = 25 ext{ cm}$. (Ignoring the required negative sign).If the X-intercept is $0.04 ext{ cm}^{-1}$ (measured distance along the axis), the actual coordinate point is $(-0.04, 0)$. Therefore: $mathbf{-0.04 = -1/f}$, yielding $f = 25 ext{ cm}$.
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