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 11.
Q31 (2026)
In parallax method for the determination of focal length of a concave mirror, the object should always be placed :
- between the focus (F) and the centre of curvature (C) of the mirror ONLY
- at any point beyond the focus (F) of the mirror
- beyond the centre of the curvature (C) of the mirror ONLY
- between the pole (P) and the focus (F) of the concave mirror ONLY
### Related Formula
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
### Core Logic
To locate an image using the optical bench parallax method, a real image must be formed in space in front of the mirror so that an optical pin can be aligned with it.
For a concave mirror, real images are formed only when the object is located beyond the principal focus ($u > f$). If placed between pole and focus, a virtual image is formed behind the mirror, preventing parallax removal using standard optical pins.
### Step 1: Final Conclusion
Hence, the object can be placed at any point beyond the focus (F) of the mirror.
### Pattern Recognition
Experimental optics rule: Parallax method requires real image formation $\implies$ object must be placed beyond focus ($u > f$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q26 (2026)
A thin prism with angle $5^{\circ}$ of refractive index 1.72 is combined with another prism of refractive index 1.9 to produce dispersion without deviation. The angle of second prism is
- $4.5^{\circ}$
- $6^{\circ}$
- $4^{\circ}$
- $5^{\circ}$
### Related Formula
$$\delta_{\text{net}} = 0$$
$$\delta_{1} + \delta_{2} = 0$$
$$(\mu_{1} - 1)A_{1} + (\mu_{2} - 1)A_{2} = 0$$
### Core Logic
For a combination of two prisms producing dispersion without deviation, the net angular deviation must be zero. The opposing orientations mean their deviations will cancel out algebraically.
### Step 1: Final Calculation
$$A_{2} = \frac{(\mu_{1} - 1)A_{1}}{(\mu_{2} - 1)}$$
$$A_{2} = \frac{(1.72 - 1)}{(1.9 - 1)} \times 5^{\circ} = \frac{0.72}{0.9} \times 5^{\circ} = 4^{\circ}$$
### Pattern Recognition
Sees: "dispersion without deviation" → net deviation is zero. Shortcut: $(\mu_{1} - 1)A_{1} = (\mu_{2} - 1)A_{2}$. Inverse relation between angle and refractive index difference.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q36 (2026)
Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are $2.4 \times 10^{8}\text{ m/s}$ and $2.7 \times 10^{8}\text{ m/s}$ respectively, then the value of critical angle is:
- $\cot^{-1}\left(\frac{3}{\sqrt{13}}\right)$
- $\sin^{-1}\left(\frac{9}{8}\right)$
- $\tan^{-1}\left(\frac{8}{\sqrt{17}}\right)$
- $\cos^{-1}\left(\frac{8}{9}\right)$
### Related Formula
$$\mu_{A} \sin c = \mu_{B} \sin 90^{\circ}$$
$$\mu = \frac{c}{v}$$
### Core Logic
For total internal reflection at the critical angle, the angle of refraction is $90^{\circ}$. Snell's law gives $\sin c = \mu_{B}/\mu_{A}$. Since refractive index is inversely proportional to velocity, this ratio is equal to the ratio of velocities $v_{A}/v_{B}$.
### Step 1: Velocity Ratio to Sine
$$\sin c = \frac{v_{A}}{v_{B}}$$
$$\sin c = \frac{2.4 \times 10^{8}}{2.7 \times 10^{8}} = \frac{24}{27} = \frac{8}{9}$$
### Step 2: Triangle Mapping to Tangent
We need the answer in terms of $\tan^{-1}$ based on the options. From a right triangle with Opposite = 8, Hypotenuse = 9:
$$\text{Adjacent} = \sqrt{9^{2} - 8^{2}} = \sqrt{81 - 64} = \sqrt{17}$$
$$\tan c = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{8}{\sqrt{17}}$$
$$c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right)$$
### Pattern Recognition
Sees: "total internal reflection" + "speeds given" → skip calculating refractive indices, directly jump to $\sin c = v_{\text{denser}} / v_{\text{rarer}}$. Use Pythagorean identity if options differ in trigonometric function.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q40 (2026)
A prism of angle $75^{\circ}$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be ____. ($\sin 15^{\circ} = 0.25$ and $\sin 25^{\circ} = 0.43$)
- $\text{between } 15^{\circ} \text{ and } 20^{\circ}$
- $15^{\circ}$
- $> 25^{\circ}$
- $< 15^{\circ}$
### Related Formula
$r_1 + r_2 = A$
Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$
Condition for TIR: $\theta > \theta_C$, where $\sin \theta_C = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}}$
### Core Logic
{{SOL_IMG1}}
Given prism angle $A = 75^{\circ}$.
For TIR at the back surface, light travels from prism ($\mu = \sqrt{3}$) to film ($\mu = 1.5$).
$$r_1 + r_2 = 75^{\circ}$$
### Step 1: Critical Angle for TIR
Applying Snell's Law for critical angle at the back surface:
$$\sqrt{3} \sin r_2 = 1.5 \sin 90^{\circ}$$
$$\sqrt{3} \sin r_2 = \frac{3}{2}$$
$$\sin r_2 = \frac{\sqrt{3}}{2} \implies r_2 = 60^{\circ}$$
To achieve TIR, we must have $r_2 \ge 60^{\circ}$.
### Step 2: Constraint on Front Surface
Since $r_1 + r_2 = 75^{\circ}$, the condition $r_2 \ge 60^{\circ}$ means:
$$r_1 \le 15^{\circ}$$
### Step 3: Constraint on Incident Angle
Applying Snell's law at the first interface (air to prism):
$$1 \cdot \sin i = \sqrt{3} \sin r_1$$
For the limit condition $r_1 = 15^{\circ}$:
$$\sin i = \sqrt{3} \sin 15^{\circ} = 1.732 \times 0.25 = 0.433$$
We know $\sin 25^{\circ} = 0.43$. Thus $0.433$ is roughly $\sin 25^{\circ}$.
So $i \le 25^{\circ}$ is required. Any angle less than $25^{\circ}$ ensures $r_1 \le 15^{\circ}$ and thus $r_2 \ge 60^{\circ}$.
### Step 4: Analyze Options
Any angle $i < 25^{\circ}$ guarantees TIR.
Option (1): between $15^{\circ}$ and $20^{\circ}$ (Valid)
Option (2): $15^{\circ}$ (Valid)
Option (4): $< 15^{\circ}$ (Valid)
Since this is an MCQ frame inherently reflecting all suitable subsets, (1), (2), and (4) are all physically sufficient.
### Pattern Recognition
In prism TIR problems, trace backward from the critical surface constraint. Maximum $r_2$ means minimum $r_1$, which dictates maximum $i$. If $i_{\text{max}} = 25^{\circ}$, any choice strictly bounded below $25^{\circ}$ is physically correct.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q47 (2026)
The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ____ cm.
### Related Formula
Magnification formula for thin lens in terms of $f$ and $u$:
$$m = \frac{f}{f + u}$$
### Core Logic
Since the sizes of the images are equal at two different object positions but the focal length doesn't change, one image must be real and the other virtual. Hence, their magnifications are equal in magnitude but opposite in sign.
$m_1 = -m_2$
### Step 1: Set Up Magnification Equations
Object distance 1: $u_1 = -8 \, \mathrm{cm}$ (virtual image case, closer to lens)
Object distance 2: $u_2 = -24 \, \mathrm{cm}$ (real image case)
Using proper sign convention:
$$\frac{f}{f + (-8)} = -\left( \frac{f}{f + (-24)} \right)$$
### Step 2: Solve for Focal Length
$$\frac{f}{f - 8} = -\frac{f}{f - 24}$$
Assuming $f \neq 0$:
$$f - 24 = -(f - 8)$$
$f - 24 = -f + 8$
$2f = 32$
$$f = 16 \, \mathrm{cm}$$
### Pattern Recognition
For a convex lens, if equal size images are formed at two different object distances $u_1$ and $u_2$, then $f = \frac{u_1 + u_2}{2}$ if both distances are strictly positive magnitude values.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
