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 12.
Q31 (2026)
The exit surface of a prism with refractive index n is coated with a material having refractive index $\frac{n}{2}$. When this prism is set for minimum angle of deviation it exactly meets the condition of critical angle. The prism angle is ____
- $60^{\circ}$
- $15^{\circ}$
- $30^{\circ}$
- $45^{\circ}$
### Related Formula
$$\text{Minimum Deviation: } r_1 = r_2 = \frac{A}{2}$$
$$\text{Critical Angle: } \sin \theta_c = \frac{\mu_2}{\mu_1}$$
### Core Logic
{{SOL_IMG1}}
For minimum deviation, $i = e$ and $r = \frac{A}{2}$.
At the exit surface, the light ray exactly meets the condition of the critical angle.
{{SOL_IMG2}}
For Total Internal Reflection (TIR) condition matching critical angle:
$r = \theta_c$
$$\sin r = \sin \theta_c$$
$$\sin r = \frac{n/2}{n} = \frac{1}{2}$$
### Step 1: Solve for Prism Angle
Since $\sin r = \frac{1}{2}$, we have $r = 30^{\circ}$.
Since $r = \frac{A}{2}$ at minimum deviation:
$$\frac{A}{2} = 30^{\circ}$$
$A = 60^{\circ}$
### Pattern Recognition
Critical angle at an interface is always $\sin^{-1}(\mu_{\text{rare}}/\mu_{\text{dense}})$. If $\mu_{\text{rare}}$ is exactly half of $\mu_{\text{dense}}$, the critical angle is exactly $30^\circ$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q33 (2026)
In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is $(5)^{k}$. The value of k is ____
- $2$
- $5$
- $3.5$
- $4$
### Related Formula
$$M = \frac{\ell}{f_0} \cdot \frac{D}{f_e}$$
### Core Logic
Given parameters:
Focal length of objective, $f_0 = 2 \text{ cm}$
Focal length of eyepiece, $f_e = 5 \text{ cm}$
Tube length, $\ell = 10 \text{ cm}$
Least distance of distinct vision, $D = 25 \text{ cm}$
### Step 1: Calculate Total Magnification
For normal adjustment (final image at infinity), the magnification formula is:
$$M = \left( \frac{\ell}{f_0} \right) \left( \frac{D}{f_e} \right)$$
$$M = \left( \frac{10}{2} \right) \left( \frac{25}{5} \right) = 5 \times 5 = 25$$
Since $M = 25 = 5^2$, and the problem states $M = (5)^k$, we compare to find $k=2$.
### Pattern Recognition
For normal adjustment of a compound microscope, $M = m_o \times m_e \approx (L/f_o) \times (D/f_e)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q28 (2026)
Five persons $P_{1}$ , $P_{2}$ , $P_{3}$ , $P_{4}$ and $P_{5}$ recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement
- $\text{Readings recorded by all persons are correct}$
- $\text{Reading recorded by } P_{3} \text{ persons are incorrect}$
- $\text{Reading recorded by } P_{3} \text{ and } P_{2} \text{ persons are incorrect}$
- $\text{Reading recorded by } P_{4} \text{ and } P_{5} \text{ persons are incorrect}$
### Related Formula
$$P = \frac{100}{f \text{ (in cm)}}$$
### Core Logic
Given Power $P = +5\mathrm{D}$
$$\frac{1}{f} = 5 \Rightarrow f = 20 \text{ cm}$$
If the object is between $f$ and $2f$ (between $20\text{cm}$ and $40\text{cm}$), the image must be formed beyond $2f$ (i.e., $v > 40\text{cm}$) and it is magnified.
### Step 1: Check P3 Reading
For $P_3$, $u = 35\text{cm}$ (which is between $20\text{cm}$ and $40\text{cm}$). Thus, the image distance $v$ must be greater than $40\text{cm}$.
However, the recorded reading is $v = 37\text{cm}$, which is incorrect.
### Pattern Recognition
For a convex lens, if $u$ lies between $f$ and $2f$, $v$ must be greater than $2f$. A simple boundary check on the $2f$ mark ($40\text{cm}$) reveals faulty readings instantly without solving the lens formula.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q45 (2026)
Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ____ cm.
- $- 15/2$
- $- 10$
- $- 20$
- $- 15$
### Related Formula
$$m = -\frac{v}{u}$$
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
### Core Logic
For a real image, magnification $m$ is negative:
$m = -3$
$$-3 = -\frac{v}{u} \implies v = 3u$$
Given the distance between object and image is $40\text{ cm}$:
For real images formed by a concave mirror, both $u$ and $v$ are on the same side. The distance between them is $|v| - |u| = 40\text{ cm}$.
### Step 1: Determine Object and Image Distances
$3|u| - |u| = 40$
$$2|u| = 40 \implies |u| = 20 \text{ cm}$$
$$|v| = 60 \text{ cm}$$
Applying sign convention:
$$u = -20 \text{ cm}, \quad v = -60 \text{ cm}$$
### Step 2: Compute Focal Length
$$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20}$$
$$\frac{1}{f} = \frac{-1 - 3}{60} = -\frac{4}{60}$$
$$f = -15 \text{ cm}$$
### Pattern Recognition
Real magnified image means concave mirror. The separation is $(m-1)|u|$. Therefore $|u| = \Delta / (m-1)$. Then strictly apply $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with signs.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q36 (2026)
The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ____.
- $5:2$
- $5:12$
- $12:5$
- $2:5$
### Related Formula
$$P = \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
### Core Logic
Equate the magnitudes of the power of the two lenses using the Lens Maker's formula.
{{SOL_IMG1}}
### Step 1: Power of Biconvex Lens
For a biconvex lens ($\mu = 1.5$), let the front radius be $R_1$ and back radius be $R_2$. By sign convention, $R_1 > 0$ and $R_2 < 0$.
$$|P_A| = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{-R_2}\right) = 0.5 \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
### Step 2: Power of Plano-Concave Lens
For a plano-concave lens ($\mu' = 1.7$), the flat surface has radius $\infty$. The curved surface matches the back surface of the biconvex lens, so its radius is $R_2$.
$$|P_B| = (1.7 - 1)\left| \frac{1}{-R_2} - \frac{1}{\infty} \right| = \frac{0.7}{R_2}$$
### Step 3: Equating Powers
Given $|P_A| = |P_B|$:
$$0.5 \left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$$
$$\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2}$$
$$\frac{0.5}{R_1} = \frac{0.2}{R_2}$$
### Step 4: Finding the Ratio
$$\frac{5}{R_1} = \frac{2}{R_2}$$
$$\frac{R_1}{R_2} = \frac{5}{2}$$
### Pattern Recognition
Matching curvatures means treating the absolute value of the radius as the same variable $R_2$ across the two equations. Always use magnitude equations when explicitly stated 'magnitudes of power are same'.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
