Welcome, aspiring physicists, to a deep dive into the fundamental equations that govern the behavior of light when it interacts with mirrors and lenses. These formulae โ the mirror formula, the lens formula, and the magnification formulae โ are the backbone of Ray Optics. Mastering them isn't just about memorizing equations; it's about understanding the underlying principles, applying correct sign conventions, and interpreting the results to predict how images are formed. This section will equip you with a robust understanding, starting from the very basics and building up to the nuances required for JEE Advanced.
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1. The Cornerstone: Cartesian Sign Convention
Before we write down any formula, we *must* establish a consistent system for measuring distances and heights. This is where the
Cartesian Sign Convention comes in. It's an absolute necessity for correctly applying mirror and lens formulae. Without it, your calculations will be incorrect, even if you know the formulae.
Imagine a standard Cartesian coordinate system.
- Origin: For mirrors, the pole (P) is taken as the origin. For lenses, the optical center (O) is the origin. All distances are measured from this point.
- Principal Axis: This is the x-axis.
- Direction of Incident Light: This is crucial! We usually assume light travels from left to right. All distances measured *against* the direction of incident light are taken as negative. All distances measured *along* the direction of incident light are taken as positive.
- Heights: Heights measured *upwards* from the principal axis are positive. Heights measured *downwards* from the principal axis are negative.
Let's summarize this in a structured way:
Parameter |
Sign Convention |
Interpretation |
|---|
Object Distance (u) |
Almost always negative for real objects (placed in front of the mirror/lens), as light travels from the object towards the optical element (from left to right), and 'u' is measured against this direction. Can be positive for virtual objects (e.g., when light from one element converges towards another). |
Distance of the object from the pole/optical center. |
Image Distance (v) |
Positive if the image is formed on the same side as incident light (e.g., real image by concave mirror, virtual image by convex lens). Negative if formed on the opposite side (e.g., virtual image by concave mirror, real image by convex lens). |
Distance of the image from the pole/optical center. A positive 'v' usually means a real image (unless it's a virtual object), a negative 'v' usually means a virtual image. |
Focal Length (f) |
Negative for concave mirror / convex lens (converging elements). Positive for convex mirror / concave lens (diverging elements). |
Distance of the principal focus from the pole/optical center. The sign indicates the nature of the optical element. |
Radius of Curvature (R) |
Negative for concave mirror (center of curvature on incident side). Positive for convex mirror (center of curvature on refracted/reflected side). |
Radius of the sphere of which the mirror/lens surface is a part. Related to 'f' by R = 2f (for mirrors) or part of lens maker's formula. |
Object Height (h) |
Positive if placed upright above the principal axis. |
Height of the object perpendicular to the principal axis. |
Image Height (h') |
Positive if erect (upright) with respect to the principal axis. Negative if inverted. |
Height of the image perpendicular to the principal axis. |
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2. The Mirror Formula: Unveiling Image Formation by Reflection
The mirror formula relates the object distance (`u`), image distance (`v`), and focal length (`f`) for spherical mirrors. It's a fundamental relationship derived from similar triangles in ray diagrams.
The Mirror Formula:
$$ frac{1}{f} = frac{1}{v} + frac{1}{u} $$
Where:
* `f` is the focal length of the mirror. Remember, `f = R/2`.
* `v` is the image distance from the pole.
* `u` is the object distance from the pole.
Key Points for Mirrors:
* For a
concave mirror (converging mirror), `f` is
negative.
* For a
convex mirror (diverging mirror), `f` is
positive.
*
Real images are formed in front of the mirror (on the same side as the object), so `v` will be
negative.
*
Virtual images are formed behind the mirror, so `v` will be
positive.
Example 1: Concave Mirror - Real Image
A concave mirror has a focal length of 20 cm. An object is placed at a distance of 30 cm from the mirror. Find the position and nature of the image.
Solution:
1.
Given:
* Focal length, `f = -20 cm` (concave mirror, so negative by sign convention).
* Object distance, `u = -30 cm` (object placed in front, against incident light).
2.
Mirror Formula:
$$ frac{1}{f} = frac{1}{v} + frac{1}{u} $$
3.
Substitute values:
$$ frac{1}{-20} = frac{1}{v} + frac{1}{-30} $$
$$ -frac{1}{20} = frac{1}{v} - frac{1}{30} $$
4.
Solve for `v`:
$$ frac{1}{v} = -frac{1}{20} + frac{1}{30} $$
$$ frac{1}{v} = frac{-3 + 2}{60} $$
$$ frac{1}{v} = -frac{1}{60} $$
$$ v = -60 ext{ cm} $$
5.
Interpretation:
* The negative sign for `v` indicates that the image is formed
in front of the mirror.
* Since it's in front, the image is
real.
Example 2: Convex Mirror - Virtual Image
An object is placed 15 cm in front of a convex mirror of focal length 10 cm. Determine the position and nature of the image.
Solution:
1.
Given:
* Object distance, `u = -15 cm`.
* Focal length, `f = +10 cm` (convex mirror, so positive).
2.
Mirror Formula:
$$ frac{1}{f} = frac{1}{v} + frac{1}{u} $$
3.
Substitute values:
$$ frac{1}{10} = frac{1}{v} + frac{1}{-15} $$
$$ frac{1}{10} = frac{1}{v} - frac{1}{15} $$
4.
Solve for `v`:
$$ frac{1}{v} = frac{1}{10} + frac{1}{15} $$
$$ frac{1}{v} = frac{3 + 2}{30} $$
$$ frac{1}{v} = frac{5}{30} = frac{1}{6} $$
$$ v = +6 ext{ cm} $$
5.
Interpretation:
* The positive sign for `v` indicates that the image is formed
behind the mirror.
* Since it's behind the mirror, the image is
virtual.
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3. Magnification for Mirrors
Magnification (`m`) tells us how much larger or smaller the image is compared to the object, and whether it's erect or inverted. For mirrors, we primarily deal with
lateral (or transverse) magnification, which is the ratio of the height of the image to the height of the object.
Lateral Magnification Formula (Mirrors):
$$ m = frac{ ext{Height of Image (h')}}{ ext{Height of Object (h)}} = -frac{v}{u} $$
Derivation Insight (using similar triangles):
Consider a ray from the top of the object striking the pole of a mirror and reflecting. The angle of incidence equals the angle of reflection. The triangles formed by the object, its image, and the principal axis, with the pole as a common vertex, are similar. This similarity leads directly to `h'/h = -v/u`.
Interpreting Magnification (`m`):
*
Sign of `m`:
* If `m` is
positive, the image is
erect (upright) relative to the object.
* If `m` is
negative, the image is
inverted.
*
Magnitude of `m` (`|m|`):
* If `|m| > 1`, the image is
enlarged.
* If `|m| < 1`, the image is
diminished.
* If `|m| = 1`, the image is of the
same size as the object.
Connecting to Example 1 (Concave Mirror):
From Example 1, `u = -30 cm` and `v = -60 cm`.
$$ m = -frac{v}{u} = -frac{-60}{-30} = -2 $$
Interpretation: `m = -2` means the image is
inverted (due to negative sign) and
twice the size of the object (`|m| = 2`).
Connecting to Example 2 (Convex Mirror):
From Example 2, `u = -15 cm` and `v = +6 cm`.
$$ m = -frac{v}{u} = -frac{+6}{-15} = frac{6}{15} = +frac{2}{5} = +0.4 $$
Interpretation: `m = +0.4` means the image is
erect (positive sign) and
0.4 times the size of the object (`|m| < 1`), i.e., diminished.
JEE Advanced Tip - Other Magnifications:
*
Axial (Longitudinal) Magnification (`m_L`): For small objects placed along the principal axis, `m_L = - (v/u)^2`. This is for object length along the axis.
*
Areal Magnification (`m_A`): For 2D objects, `m_A = m_L * m_T = m^2`.
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4. The Lens Formula: Unveiling Image Formation by Refraction
Similar to mirrors, lenses also have a formula relating object distance (`u`), image distance (`v`), and focal length (`f`). However, there's a crucial sign difference due to refraction.
The Lens Formula:
$$ frac{1}{f} = frac{1}{v} - frac{1}{u} $$
Key Points for Lenses:
* For a
convex lens (converging lens), `f` is
positive.
* For a
concave lens (diverging lens), `f` is
negative.
*
Real images are formed on the opposite side of the object (on the right, if light comes from left), so `v` will be
positive.
*
Virtual images are formed on the same side as the object (on the left), so `v` will be
negative.
JEE Focus - Lens Maker's Formula:
The focal length of a lens is determined by its refractive index and the radii of curvature of its surfaces.
$$ frac{1}{f} = (n_{rel} - 1) left( frac{1}{R_1} - frac{1}{R_2}
ight) $$
Where `n_rel = n_lens / n_medium` (refractive index of lens material relative to the surrounding medium). `R1` and `R2` are the radii of curvature of the two surfaces, with their signs determined by the convention: radius is positive if its center of curvature is on the side of refracted light (e.g., for a convex surface refracting light from left to right, R is positive if its center is to the right).
Example 3: Convex Lens - Real Image
A convex lens has a focal length of 15 cm. An object is placed at a distance of 25 cm from the lens. Find the position and nature of the image.
Solution:
1.
Given:
* Focal length, `f = +15 cm` (convex lens, so positive).
* Object distance, `u = -25 cm` (object placed in front, against incident light).
2.
Lens Formula:
$$ frac{1}{f} = frac{1}{v} - frac{1}{u} $$
3.
Substitute values:
$$ frac{1}{15} = frac{1}{v} - frac{1}{-25} $$
$$ frac{1}{15} = frac{1}{v} + frac{1}{25} $$
4.
Solve for `v`:
$$ frac{1}{v} = frac{1}{15} - frac{1}{25} $$
$$ frac{1}{v} = frac{5 - 3}{75} $$
$$ frac{1}{v} = frac{2}{75} $$
$$ v = frac{75}{2} = +37.5 ext{ cm} $$
5.
Interpretation:
* The positive sign for `v` indicates that the image is formed
on the opposite side of the object (to the right).
* Since it's on the opposite side, the image is
real.
Example 4: Concave Lens - Virtual Image
An object is placed 10 cm in front of a concave lens of focal length 15 cm. Determine the position and nature of the image.
Solution:
1.
Given:
* Object distance, `u = -10 cm`.
* Focal length, `f = -15 cm` (concave lens, so negative).
2.
Lens Formula:
$$ frac{1}{f} = frac{1}{v} - frac{1}{u} $$
3.
Substitute values:
$$ frac{1}{-15} = frac{1}{v} - frac{1}{-10} $$
$$ -frac{1}{15} = frac{1}{v} + frac{1}{10} $$
4.
Solve for `v`:
$$ frac{1}{v} = -frac{1}{15} - frac{1}{10} $$
$$ frac{1}{v} = frac{-2 - 3}{30} $$
$$ frac{1}{v} = -frac{5}{30} = -frac{1}{6} $$
$$ v = -6 ext{ cm} $$
5.
Interpretation:
* The negative sign for `v` indicates that the image is formed
on the same side as the object (to the left).
* Since it's on the same side, the image is
virtual.
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5. Magnification for Lenses
Similar to mirrors, the lateral magnification (`m`) for lenses is defined as the ratio of image height to object height.
Lateral Magnification Formula (Lenses):
$$ m = frac{ ext{Height of Image (h')}}{ ext{Height of Object (h)}} = frac{v}{u} $$
Derivation Insight (using similar triangles):
For lenses, consider a ray from the top of the object passing straight through the optical center (un-deviated). The triangles formed by the object, its image, and the principal axis, with the optical center as a common vertex, are similar. This similarity leads directly to `h'/h = v/u`. Note the absence of the negative sign compared to mirrors.
Interpreting Magnification (`m`):
The interpretation rules for the sign and magnitude of `m` are identical to those for mirrors:
*
Sign of `m`: Positive for erect, negative for inverted.
*
Magnitude of `m` (`|m|`): `|m| > 1` (enlarged), `|m| < 1` (diminished), `|m| = 1` (same size).
Connecting to Example 3 (Convex Lens):
From Example 3, `u = -25 cm` and `v = +37.5 cm`.
$$ m = frac{v}{u} = frac{+37.5}{-25} = -1.5 $$
Interpretation: `m = -1.5` means the image is
inverted and
1.5 times the size of the object.
Connecting to Example 4 (Concave Lens):
From Example 4, `u = -10 cm` and `v = -6 cm`.
$$ m = frac{v}{u} = frac{-6}{-10} = +0.6 $$
Interpretation: `m = +0.6` means the image is
erect and
0.6 times the size of the object (diminished).
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6. Power of a Lens and Combination of Lenses
Power of a Lens (P):
The power of a lens is a measure of its ability to converge or diverge light rays. It is defined as the reciprocal of its focal length.
$$ P = frac{1}{f} $$
* The SI unit for power is the
dioptre (D), where 1 D = 1 mโปยน.
* For `P` to be in dioptres, `f` must be in meters.
*
Convex lenses have positive power (converging).
*
Concave lenses have negative power (diverging).
Combination of Thin Lenses in Contact:
When several thin lenses are placed in contact, the equivalent focal length (`F_eq`) and equivalent power (`P_eq`) are given by:
$$ frac{1}{F_{eq}} = frac{1}{f_1} + frac{1}{f_2} + frac{1}{f_3} + dots $$
Or in terms of power:
$$ P_{eq} = P_1 + P_2 + P_3 + dots $$
The total magnification for a combination of lenses is the product of individual magnifications: `m_total = m1 * m2 * m3 * ...`
Example 5: Combined Lenses
Two thin lenses, one convex with focal length +20 cm and another concave with focal length -10 cm, are placed in contact. Find the focal length and power of the combination. If an object of height 2 cm is placed 30 cm from this combination, find the position, nature, and height of the final image.
Solution:
1.
Given:
* `f1 = +20 cm` (convex lens)
* `f2 = -10 cm` (concave lens)
* `h = 2 cm`
* `u_combination = -30 cm`
2.
Equivalent Focal Length (`F_eq`):
$$ frac{1}{F_{eq}} = frac{1}{f_1} + frac{1}{f_2} = frac{1}{+20} + frac{1}{-10} = frac{1}{20} - frac{2}{20} = -frac{1}{20} $$
$$ F_{eq} = -20 ext{ cm} $$
The combination acts as a
concave lens (diverging), as its focal length is negative.
3.
Equivalent Power (`P_eq`):
$$ P_1 = frac{1}{0.20 ext{ m}} = +5 ext{ D} $$
$$ P_2 = frac{1}{-0.10 ext{ m}} = -10 ext{ D} $$
$$ P_{eq} = P_1 + P_2 = +5 ext{ D} - 10 ext{ D} = -5 ext{ D} $$
4.
Image formation by the combination:
Using the lens formula for the equivalent lens:
$$ frac{1}{F_{eq}} = frac{1}{v} - frac{1}{u} $$
$$ frac{1}{-20} = frac{1}{v} - frac{1}{-30} $$
$$ -frac{1}{20} = frac{1}{v} + frac{1}{30} $$
$$ frac{1}{v} = -frac{1}{20} - frac{1}{30} = frac{-3 - 2}{60} = -frac{5}{60} = -frac{1}{12} $$
$$ v = -12 ext{ cm} $$
The negative sign for `v` indicates the image is formed
on the same side as the object (virtual).
5.
Magnification and Image Height:
$$ m = frac{v}{u} = frac{-12}{-30} = +frac{2}{5} = +0.4 $$
The positive sign for `m` indicates the image is
erect.
$$ h' = m imes h = 0.4 imes 2 ext{ cm} = 0.8 ext{ cm} $$
The image is
diminished (0.8 cm height).
Final Interpretation: The combination forms a virtual, erect, and diminished image 12 cm from the lenses, on the same side as the object, with a height of 0.8 cm.
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7. JEE Advanced Corner: Beyond Basic Formulae
*
Newton's Formula: For mirrors and lenses, if object distance `x` is measured from the principal focus (instead of pole/optical center) and image distance `x'` is also measured from the focus, then:
* For mirrors: `xx' = f^2` (Here, x and x' are distances from focus, not coordinates).
* For lenses: `xx' = f^2`
This formula is particularly useful when questions provide distances relative to the focal point.
*
Velocity of Image: If an object is moving, its image will also move. The velocity of the image can be found by differentiating the mirror/lens formula with respect to time.
* For mirrors (`1/v + 1/u = 1/f`): `(-1/v^2)(dv/dt) + (-1/u^2)(du/dt) = 0`
$$ frac{dv}{dt} = -left(frac{v}{u}
ight)^2 frac{du}{dt} $$
Where `dv/dt` is velocity of image, `du/dt` is velocity of object.
* For lenses (`1/v - 1/u = 1/f`): `(-1/v^2)(dv/dt) - (-1/u^2)(du/dt) = 0`
$$ frac{dv}{dt} = left(frac{v}{u}
ight)^2 frac{du}{dt} $$
Remember to use proper sign conventions for velocities (e.g., if object moves towards the mirror/lens, `du/dt` might be negative if `u` is inherently negative and its magnitude is decreasing).
*
Displacement Method for Convex Lens: This is an experimental method to find the focal length of a convex lens. If for a fixed distance `D` between an object and a screen, a convex lens produces a real image on the screen for two positions (let the distance between these two positions be `x`), then `f = (D^2 - x^2) / (4D)`. This method applies when `D > 4f`.
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By thoroughly understanding the sign conventions, the formulae themselves, and their interpretations, you'll be well-equipped to tackle a wide range of problems involving mirrors and lenses, from basic board-level questions to complex JEE challenges. Practice is key, so apply these concepts diligently to various problems!