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 5.
Q909 (2025)
When an object is placed $40\mathrm{~cm}$ away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
- 40 cm away from the mirror.
- 80 cm away from the mirror.
- 20 cm towards the mirror.
- 20 cm away from the mirror.
### Related Formula
Magnification formula for a spherical mirror in terms of focal length $f$ and object distance $u$:
$$m = \frac{f}{f - u}$$
### Core Logic
**Case 1:** Given $u_1 = -40\mathrm{~cm}$ and $m_1 = \frac{1}{2}$:
$$\frac{1}{2} = \frac{f}{f - (-40)} \implies f + 40 = 2f \implies f = +40\mathrm{~cm}$$
Since $f > 0$, the optical element is a **convex mirror** forming a virtual, erect, and diminished image ($0 < m < 1$).
### Step 1: Calculate New Object Position
**Case 2:** To achieve a magnification of $m_2 = \frac{1}{3}$:
$$\frac{1}{3} = \frac{40}{40 - u_2} \implies 40 - u_2 = 120 \implies u_2 = -80\mathrm{~cm}$$
### Step 2: Determine Distance Shift
* Initial object position: $u_1 = -40\mathrm{~cm}$
* Final object position: $u_2 = -80\mathrm{~cm}$
$$\text{Shift} = |u_2| - |u_1| = 80 - 40 = 40\mathrm{~cm}\text{ away from the mirror}$$
### Pattern Recognition
For a convex mirror, images are always virtual, erect, and diminished. As the object is shifted farther away from the mirror ($u \to -\infty$), the magnification decreases toward zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q923 (2025)
Distance between object and its image (magnified by $-\frac{1}{3}$) is $30\text{ cm}$. The focal length of the mirror used is $\left(\frac{x}{4}\right)\text{ cm}$, where magnitude of value of $x$ is
### Related Formula
Magnification relation for spherical mirrors:
$$m = -\frac{v}{u}$$
Mirror equation:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
### Core Logic
Given $m = -\frac{1}{3}$:
$$-\frac{v}{u} = -\frac{1}{3} implies u = 3v$$
Since magnification is negative, a real inverted image is formed on the same side as the object in a concave mirror layout configuration.
{{SOL_IMG1}}
### Step 1: Formulate Position Distances
The distance between the object and image is given as $30\mathrm{~cm}$:
$$|u| - |v| = 30 implies 3v - v = 30 implies 2v = 30 implies v = 15\mathrm{~cm}$$
This gives $u = 3(15) = 45\mathrm{~cm}$.
### Step 2: Solve for Focal Length and x
Apply standard sign conventions ($u = -45\mathrm{~cm}$, $v = -15\mathrm{~cm}$):
$$\frac{1}{f} = -\frac{1}{15} - \frac{1}{45} = \frac{-3 - 1}{45} = -\frac{4}{45}$$
$$|f| = \frac{45}{4}\mathrm{~cm}$$
Matching this with the prompt pattern form $\frac{x}{4}$ yields:
$x = 45$
### Pattern Recognition
Real inverted diminished images ($|m| < 1$) mean the image forms closer to the mirror surface than the object, situated between the focal center $F$ and center of curvature $C$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q9 (2025)
A mirror is used to produce an image with magnification of $\frac{1}{4}$
If the distance between object and its image is 40 cm, then the focal length of the mirror is [cite: 113, 114, 116]
- 10 cm [cite: 117]
- 12.7 cm [cite: 119]
- 10.7 cm [cite: 118]
- 15 cm [cite: 119]
### Related Formula
$$m = -\frac{v}{u}$$ [cite: 700]
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ [cite: 707]
### Core Logic
Given magnification magnitude $|m| = \frac{1}{4}$[cite: 113]. Assuming a real image formed by a concave mirror: [cite: 701]
$$\frac{v}{u} = \frac{1}{4} \implies u = 4v$$ [cite: 701]
The distance between the object and the image is given as $40\ \text{cm}$ [cite: 114, 116]:
$$u - v = 40 \implies 4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3}\ \text{cm}$$
$$u = 4 \times \frac{40}{3} = \frac{160}{3}\ \text{cm}$$
Applying mirror sign conventions ($u = -\frac{160}{3}$, $v = -\frac{40}{3}$): [cite: 701, 704]
$$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = -\frac{12 + 3}{160} = -\frac{15}{160}$$
$$f = -\frac{160}{15} \approx -10.67\ \text{cm}$$ [cite: 712]
Rounding to the matching options choice gives $10.7\ \text{cm}$[cite: 118, 712].
### Pattern Recognition
Pay attention to sign conventions in mirror systems. A smaller real image formed by a concave mirror sits between the focus and center of curvature, resulting in $u > v$ configurations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q14 (2025)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Refractive index of glass is higher than that of air. [cite: 130]
Reason (R): Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index. [cite: 131]
In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 132]
- (A) is not correct but (R) is correct [cite: 133]
- Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 134]
- (A) is correct but (R) is not correct [cite: 135]
- Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 136]
### Core Logic
Refractive index represents the ratio of the speed of light in vacuum to its speed in a given medium[cite: 719]. Glass slows light down more than air does, hence $\mu_{\text{glass}} \approx 1.5 > \mu_{\text{air}} \approx 1.0$, which makes Assertion (A) correct[cite: 130]. However, optical density is defined by a medium's capacity to refract light and is completely conceptually distinct from inertial mass density (mass per unit volume)[cite: 719]. For example, turpentine has a lower mass density than water but possesses a higher optical density and refractive index. Therefore, Reason (R) is fundamentally incorrect[cite: 722].
### Pattern Recognition
Optical density vs mass density is a signature conceptual trick in refraction theory[cite: 719]. They share the word 'density' but have entirely different physical meanings and no fixed mathematical proportionality[cite: 719, 722].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q20 (2025)
A transparent block A having refractive index $\mu=1.25$ is surrounded by another medium of refractive index $\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle $\theta$ as shown in figure. What is the maximum value of $\theta$ for which light suffers total internal reflection at the top surface of the block?
{{IMG1}} [cite: 169, 170, 171]
- $\tan^{-1}(4/3)$ [cite: 175]
- $\tan^{-1}(3/4)$ [cite: 176]
- $\sin^{-1}(3/4)$ [cite: 177]
- $\cos^{-1}(3/4)$ [cite: 178]
### Related Formula
$$\sin\theta_c = \frac{\mu_1}{\mu_2}$$ [cite: 810]
$$\mu_1 \sin\theta = \mu_2 \sin r$$ [cite: 807]
### Core Logic
From the boundary geometry at the top interface, the angle of refraction $r$ at the first surface satisfies: [cite: 806]
$$r + \theta_c = 90^{\circ} \implies r = 90^{\circ} - \theta_c$$ [cite: 806]
Applying Snell's law at the first entry interface: [cite: 170, 807]
$$\mu_1 \sin\theta = \mu_2 \sin r = \mu_2 \sin(90^{\circ} - \theta_c) = \mu_2 \cos\theta_c$$ [cite: 172, 173, 807, 809]
Since $\sin\theta_c = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{4}{5}$, we have $\cos\theta_c = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$[cite: 169, 810, 811]. Substituting back into the expression: [cite: 811]
$$1.0 \cdot \sin\theta = 1.25 \times \frac{3}{5} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$$ [cite: 169, 811]
$$\theta = \sin^{-1}\left(\frac{3}{4}
ight)$$ [cite: 811]
### Pattern Recognition
Maximum angle at the entry face ensures minimum angle of incidence at the subsequent wall[cite: 806, 807]. Setting that exact internal angle equal to the critical threshold condition values solves for the operational scanning range edge directly[cite: 808].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
