Ray Optics and Optical Instruments Previous Year Questions — JEE Main Physics
63 past-year Ray Optics and Optical Instruments questions from JEE Main (Physics) — page 6.
Q14 (2025)
A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is $2 \, cm \times 2 \, cm$ and the area of the landscape photographed is $400 \, km^2$ . The focal length of the lens in the drone camera is:
- 1.8 cm
- 2.8 cm
- 2.5 cm
- 0.9 cm
### Related Formula
Areal Magnification:
$$m^2 = \frac{A_{\text{image}}}{A_{\text{object}}} = \left(\frac{f}{f+u}\right)^2 \approx \left(\frac{f}{u}\right)^2$$
since object distance $u = -18\ \mathrm{km}$ is vastly larger than $f$.
### Core Logic
Given parameters: {{SOL_IMG1}}
- Object height distance, $H = 18\ \mathrm{km} = 18 \times 10^3\ \mathrm{m}$
- Film size area, $A_{\text{image}} = 2\ \mathrm{cm} \times 2\ \mathrm{cm} = 4\ \mathrm{cm^2} = 4 \times 10^{-4}\ \mathrm{m^2}$
- Landscape area, $A_{\text{object}} = 400\ \mathrm{km^2} = 400 \times 10^6\ \mathrm{m^2}$
Linear magnification factor:
$$\frac{y}{x} = \sqrt{\frac{A_{\text{image}}}{A_{\text{object}}}} = \sqrt{\frac{4 \times 10^{-4}}{400 \times 10^6}} = \sqrt{10^{-12}} = 10^{-6}$$
Using the simple pinhole/thin lens perspective ratio:
$$\frac{f}{H} = 10^{-6} \implies f = 18 \times 10^3 \times 10^{-6} = 18 \times 10^{-3}\ \mathrm{m} = 1.8\ \mathrm{cm}$$
### Pattern Recognition
For aerial satellite imaging contexts where $u \gg f$, linear sizing scales directly as $\frac{\text{film side}}{\text{ground side}} = \frac{f}{H}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1137 (2025)
A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-
- 0.15 m
- 0.10 m
- 0.20 m
- 0.25 m
### Related Formula
The net focal power of a silvered tracking lens system is given by:
$$P = 2P_{L} + P_{M}$$
$$\frac{1}{f} = \frac{2}{f_{L}} + \frac{1}{f_{M}}$$
### Core Logic
As shown in diagram {{SOL_IMG1}}, the plane flat side boundary interface has an infinite radius of curvature ($R_{2} = \infty$), meaning its mirror focal component is $f_{M} = \infty \implies P_{M} = 0$. The power depends entirely on the refraction step:
$$\frac{1}{f} = \frac{2}{f_{L}} $$
### Step 1: Lens Maker Formulation
Find the focal expression of the immersed lens element [cite: 91, 679]:
$$\frac{1}{f_{L}} = \left(\frac{\mu_{\text{glass}}}{\mu_{\text{liquid}}} - 1
ight)\left(\frac{1}{R}
ight) = \left(\frac{1.5}{1.2} - 1
ight)\frac{1}{R} = \frac{0.3}{1.2}\frac{1}{R} = \frac{1}{4R}$$
Now insert this into the total system tracking balance relation :
$$\frac{1}{f} = 2 \left(\frac{1}{4R}
ight) = \frac{1}{2R} $$
Given the final effective concave configuration matches $f = 0.2\text{ m}$ :
$$\frac{1}{0.2} = \frac{1}{2R} \implies 2R = 0.2 \implies R = 0.10\text{ m}$$
### Pattern Recognition
Silvering a plano-flat back boundary means light traverses the initial curved face interface exactly twice, mapping to $R = 2 \cdot f \cdot (\mu_{\text{rel}} - 1)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1140 (2025)
A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $f_{2}$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of $f_{1}$ and $f_{2}$ will be :-
- 3:5
- 1:3
- 1:2
- 2:3
### Related Formula
Lens Maker's Formula for a lens in a surrounding medium of refractive index $\mu_{m}$ is:
$$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{m}} - 1\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$$
### Core Logic
For a plano-convex configuration, the flat side has an infinite radius of curvature ($R_{2} = \infty \implies \frac{1}{R_{2}} = 0$).
### Step 1: Evaluating Respective Focal Scales
For the first lens setup in air ($\mu_{m} = 1$) :
$$\frac{1}{f_{1}} = (1.5 - 1)\left(\frac{1}{2} - 0\right) = 0.5 \times \frac{1}{2} = \frac{1}{4} \implies f_{1} = 4\text{ cm}$$
For the second lens setup immersed inside fluid ($\mu_{m} = 1.2$) :
$$\frac{1}{f_{2}} = \left(\frac{1.5}{1.2} - 1\right)\left(\frac{1}{3} - 0\right) = (1.25 - 1)\frac{1}{3} = \frac{0.25}{3} = \frac{1}{12} \implies f_{2} = 12\text{ cm}$$
Taking their direct ratio :
$$f_{1} : f_{2} = 4 : 12 = 1 : 3$$
### Pattern Recognition
Always separate the refractive index multiplier from the geometric shape factor. This lets you calculate each change independently before taking the final ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q2 (2025)
What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D ? ['D' stands for dioptre]
- 0.04
- 0.40
- 0.1
- 0.01
### Related Formula
The relationship between optical power $P$ and focal length $F$ is given by:
$$F = \frac{1}{P}$$
Relative decrease in focal length is defined as:
$$\frac{\Delta F}{F} = \frac{F - F'}{F}$$
### Core Logic
Given initial power $P = 2.5\text{ D}$[cite: 19, 600]. After an increase of $0.1\text{ D}$, the new power is[cite: 19, 603]:
$$P' = 2.5 + 0.1 = 2.6\text{ D}$$
### Step 1: Calculate Focal Length Change
Find the initial and final focal lengths [cite: 602, 604]:
$$F = \frac{1}{2.5} = \frac{2}{5}$$
$$F' = \frac{1}{2.6} = \frac{5}{13}$$
Now, calculate the relative decrease:
$$\frac{F - F'}{F} = 1 - \frac{F'}{F} = 1 - \frac{P}{P'} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} = \frac{1}{26} \approx 0.04$$
### Pattern Recognition
For a small change, we can approximate using differentiation: $P = \frac{1}{F} \implies dP = -\frac{dF}{F^2} \implies \frac{dF}{F} = -\frac{dP}{P}$. Thus, the relative change magnitude is $\frac{0.1}{2.5} = \frac{1}{25} = 0.04$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q6 (2025)
In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at $13\mathrm{cm}$ from the vertex of the meniscus in A forms an image with a magnification of $-2$ then the radius of curvature of meniscus is :
- $1 \, \text{cm}$
- $\frac{1}{3}\mathrm{cm}$
- $\frac{2}{3} \mathrm{~cm}$
- $\frac{4}{3} \mathrm{~cm}$
### Related Formula
For refraction at a single spherical surface separating two mediums :
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$
Linear magnification for a spherical refracting boundary is given by:
$$m = \frac{v / n_2}{u / n_1} = \frac{v \cdot n_1}{u \cdot n_2}$
### Core Logic
Given parameters from the \text [cite: 17, 651, 654]:
* Refractive index of Medium A, $n_1 = 1.3$
* Refractive index of Medium B, $n_2 = 1.4$
* Object distance, $u = -13 \text{ cm}$
* Magnification, $m = -2$
Using the magnification formula to locate image position $v$ :
$$-2 = \frac{v \cdot 1.3}{(-13) \cdot 1.4}$$
$$-2 = \frac{1.3 \cdot v}{-18.2} \implies 1.3 v = 36.4 \implies v = 28 \text{ cm}$$
Now substitute $u = -13$, $v = 28$, $n_1 = 1.3$, $n_2 = 1.4$ into the boundary equation:
$$\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}$$
$$\frac{1}{20} + \frac{1}{10} = \frac{0.1}{R}$$
$$\frac{1 + 2}{20} = \frac{0.1}{R} \implies \frac{3}{20} = \frac{1}{10R}$$
$$30 R = 20 \implies R = \frac{2}{3} \text{ cm}$$
### Step 1: Visual Context
The visual system configuration of the refracting interface is tracked here:
{{SOL_IMG1}}
### Pattern Recognition
Keep precise track of sign conventions for single spherical surfaces. A convex meniscus towards A means the center of curvature lies inside medium B, meaning $R$ will mathematically return as a positive parameter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
