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 9.
Q1734 (2024)
In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.
The focal length of the lens is ________ $\mathrm{cm}$.
{{IMG1}}
### Related Formula
$$\text{Newton's Lens Formula: } x_1 x_2 = f^2$$
where $x_1$ and $x_2$ are object and image distances from the focal points.
### Core Logic
Since distances $x$ and $y$ are measured from the focal point (not the optical center), we use Newton's formula: $x y = f^2$.
From the graph, a prominent point on the curve is $(20, 20)$.
### Step 1: Calculate Focal Length
$$20 \times 20 = f^2$$
$$f^2 = 400 \implies f = 20 \mathrm{~cm}$$
Alternatively using standard lens formula:
Object distance from optical center $u = -(f + x)$
Image distance $v = +(f + y)$
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{f+y} - \frac{1}{-(f+x)} = \frac{1}{f}$$
If $x = y = 20 \mathrm{~cm}$:
$$\frac{1}{f+20} + \frac{1}{f+20} = \frac{1}{f}$$
$$\frac{2}{f+20} = \frac{1}{f}$$
$$2f = f + 20 \implies f = 20 \mathrm{~cm}$$
### Pattern Recognition
Whenever distances are specified relative to the focal point, Newton's formula ($xy = f^2$) instantly solves the problem.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q55 (2024)
The distance between object and its two times magnified real image as produced by a convex lens is $45 \mathrm{~cm}$. The focal length of the lens used is $\_ \_ \_ \_ \_ \_ \mathrm{cm}$
### Related Formula
$$m = \frac{v}{u}$$
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
### Core Logic
For a real image produced by a convex lens, the magnification $m$ is negative. The distance between the object and the real image is the absolute sum of their distances from the lens: $|v| + |u| = v - u = 45 \mathrm{~cm}$ (since $u$ is negative and $v$ is positive).
### Step 1: Set Up Magnification and Distance
$$\frac{v}{u} = -2$$
$$v = -2u \quad \dots (i)$$
Distance between object and image:
$$v - u = 45 \quad \dots (ii)$$
### Step 2: Solve for Object and Image Distances
Substitute (i) into (ii):
$(-2u) - u = 45$
$$-3u = 45 \Rightarrow u = -15 \mathrm{~cm}$$
Then, $v = -2(-15) = +30 \mathrm{~cm}$.
### Step 3: Calculate Focal Length
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$
$$\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$$
$$f = +10 \mathrm{~cm}$$
### Pattern Recognition
For a real image, total object-to-image distance $D = v + |u|$. Applying sign convention naturally resolves this to $D = v - u$. Two conditions (magnification + total distance) reliably solve for both $u, v$ before hitting the lens equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q52 (2024)
Light from a point source in air falls on a convex curved surface of radius $20 \text{ cm}$ and refractive index 1.5. If the source is located at $100 \text{ cm}$ from the convex surface, the image will be formed at ______ cm from the object.
### Related Formula
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
### Core Logic
For a convex refracting surface, the radius of curvature $R$ is positive if the center of curvature lies in the denser medium.
$
\mu_1 = 1 \text{ (air)}
$
$
\mu_2 = 1.5 \text{ (glass/medium)}
$
$
u = -100 \text{ cm}
$
$
R = +20 \text{ cm}
$
{{SOL_IMG1}}
### Step 1: Calculate Image Position
$$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$
$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$
$$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}$$
$$\frac{1.5}{v} = \frac{5 - 2}{200} = \frac{3}{200}$$
$$v = \frac{1.5 \times 200}{3} = 100 \text{ cm}$$
### Step 2: Distance from Object
The image is formed at $v = +100 \text{ cm}$ from the pole, which is on the other side of the surface.
The object is at $|u| = 100 \text{ cm}$ from the pole.
Distance between object and image $= |u| + v = 100 + 100 = 200 \text{ cm}$.
### Pattern Recognition
Always read the final line of optics questions carefully. The question asks for distance "from the object" not "from the surface/pole". This is a classic trap where students answer 100 instead of 200.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q34 (2024)
The refractive index of a prism with apex angle $A$ is $\cot(A/2)$. The angle of minimum deviation is :
- $\delta_{\mathrm{m}} = 180^{\circ} - A$
- $\delta_{\mathrm{m}} = 180^{\circ} - 3A$
- $\delta_{\mathrm{m}} = 180^{\circ} - 4A$
- $\delta_{\mathrm{m}} = 180^{\circ} - 2A$
### Related Formula
$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
### Core Logic
Given that the refractive index $\mu = \cot\left(\frac{A}{2}\right)$.
Substituting this into the prism formula:
$$\cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
Equating the numerators:
$$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$
We can rewrite cosine in terms of sine:
$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$
### Step 2: Solve for Deviation
Comparing the angles inside the sine functions:
$$\frac{\pi}{2} - \frac{A}{2} = \frac{A}{2} + \frac{\delta_m}{2}$$
Multiply the entire equation by 2:
$$\pi - A = A + \delta_m$$
$$\delta_m = \pi - 2A$$
Converting radians to degrees:
$$\delta_m = 180^{\circ} - 2A$$
### Pattern Recognition
Whenever refractive index $\mu = \cot(A/2)$, the relation $\sin(90^\circ - A/2)$ strictly matches the prism sine equation, meaning minimum deviation $\delta_m$ is always $180^\circ - 2A$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics And Optical Instruments
Q2091 (2026)
A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.
### Related Formula
$$\text{Magnification } m = \frac{D_{\text{out}}}{D_{\text{in}}} = \frac{f_2}{f_1}$$
### Core Logic
For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.
{{SOL_IMG1}}
By similar triangles at the focal point:
$$\tan \theta = \frac{D_1 / 2}{f_1} = \frac{D_2 / 2}{f_2}$$
### Step 1: Solving for f2
$$\frac{D_1}{f_1} = \frac{D_2}{f_2}$$
$$\frac{2}{40} = \frac{14}{f_2}$$
$$f_2 = 14 \times \frac{40}{2} = 14 \times 20 = 280\text{ mm}$$
### Pattern Recognition
A standard beam expander (Keplerian telescope design used in reverse) has $f_2/f_1 = D_2/D_1$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
