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 7.
Q13 (2025)
A concave mirror produces an image of an object such that the distance between the object and image is $20 \, \text{cm}$. If the magnification of the image is $-3'$ , then the magnitude of the radius of curvature of the mirror is:
- $3.75\mathrm{cm}$
- $30\mathrm{cm}$
- $7.5\mathrm{cm}$
- $15\mathrm{cm}$
### Related Formula
Magnification $m$ of a mirror is given by:
$$m = -\frac{v}{u}$$
The mirror equation relates focal length to distance positions:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies f = \frac{uv}{u+v}$$
Radius of curvature $R = 2f$.
### Core Logic
Given, magnification $m = -3$. This tells us the image is real and inverted[cite: 131, 757]:
$$-3 = -\frac{v}{u} \implies v = 3u$$
Since both real objects and real images lie on the same side in front of a concave mirror, $u$ and $v$ are both negative fields. The physical distance separation between them is:
$$|v| - |u| = 20 \implies 3|u| - |u| = 20 \implies 2|u| = 20 \implies |u| = 10 \text{ cm}$$
Therefore, object distance $u = -10 \text{ cm}$ and image distance $v = -30 \text{ cm}$ .
Substitute into the focal equation formula:
$$f = \frac{(-10)(-30)}{-10 - 30} = \frac{300}{-40} = -7.5 \text{ cm}$$
$$R = 2 \times |f| = 2 \times 7.5 = 15 \text{ cm}$$
### Step 1: Visual Diagram
The visual positioning profile tracking focal path boundaries is given below:
{{SOL_IMG1}}
### Pattern Recognition
A magnification of $-3$ tells you immediately that the object lies between the Focus ($F$) and Center of Curvature ($C$), while the image forms beyond $C$. This geometric layout instantly verifies that the radius value must exceed $10\text{ cm}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
Q1283 (2025)
At the interface between two materials having refractive indices $n_1$ and $n_2$ , the critical angle for reflection of an em wave is $\theta_{1C}$ . The $n_2$ material is replaced by another material having refractive index $n_3$ , such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2C}$ . If $n_3 > n_2 > n_1$ ; $\frac{n_2}{n_3} = \frac{2}{5}$ and $\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$ , then $\theta_{1C}$ is
- $\sin^{-1}\left(\frac{2}{3n_1}\right)$
- $\sin^{-1}\left(\frac{5}{6n_1}\right)$
- $\sin^{-1}\left(\frac{5}{6n_1}\right)$
- $\sin^{-1}\left(\frac{1}{3n_1}\right)$
### Core Logic
Critical angle setups dictate :
$$\sin \theta_{1C} = \frac{n_1}{n_2}, \quad \sin \theta_{2C} = \frac{n_1}{n_3}$$
Given $\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$
$$\frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2} \implies n_1 \left(\frac{2}{5} - 1\right) = \frac{n_2}{2}$$
$$\frac{n_1}{n_2} = -\frac{5}{6}$$
This produces an impossible negative value for a refractive index ratio, meaning the data contains a contradiction. This question was dropped by NTA.
Q1292 (2025)
Let $\mathbf{u}$ and $\mathbf{v}$ be the distances of the object and the image from a lens of focal length $f$ . The correct graphical representation of $\mathbf{u}$ and $\mathbf{v}$ for a convex lens when $|\mathbf{u}| > f$ , is
-
### Related Formula
$$(u - f)(v - f) = f^2$$
> **Note:** Expanding $(u - f)(v - f) = f^2$ yields $uv - f(u + v) = 0$, which simplifies directly to the magnitude form of the lens formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
---
### Core Logic
#### 1. Lens Formula and Sign Conventions
For a convex lens forming a real image ($|u| > f$):
* **Object distance ($u$):** Negative ($-u$)
* **Image distance ($v$):** Positive ($+v$)
* **Focal length ($f$):** Positive ($+f$)
Substituting these into the standard lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$$\frac{1}{v} - \left(-\frac{1}{u}\right) = \frac{1}{f}$$
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
---
#### 2. Analyzing Graphical Behavior
##### **A. The $v$ vs. $u$ Graph**
Rearranging the equation for $v$:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf} \implies v = \frac{uf}{u - f}$$
* **When $u \to f$:** $v \to \infty$ (Vertical asymptote at $u = f$)
* **When $u \to \infty$:** $v \to f$ (Horizontal asymptote at $v = f$)
* **When $u = 2f$:** $v = 2f$
This relationship represents a **rectangular hyperbola** when plotting magnitudes (or in the respective coordinate quadrants under Cartesian sign conventions).
##### **B. The $\frac{1}{v}$ vs. $\frac{1}{u}$ Graph**
Plotting the reciprocals ($\frac{1}{v}$ on the $y$-axis and $\frac{1}{u}$ on the $x$-axis):
$$\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$$
This matches the straight-line equation $y = mx + c$:
* **Slope ($m$):** $-1$
* **Intercept ($c$):** $\frac{1}{f}$
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1298 (2025)
Two light beams fall on a transparent material block at point 1 and 2 with angle $\theta_{1}$ and $\theta_{2}$ , respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given: the distance between 1 and 2, $d = 4\sqrt{3} \mathrm{~cm}$ and $\theta_{1} = \theta_{2} = \cos^{-1}\left(\frac{n_{2}}{2n_{1}}\right)$ , where refractive index of the block $n_{2} >$ refractive index of the outside medium $n_{1}$ , then the thickness of the block is ______ cm.
{{IMG1}}
### Related Formula
$$n_1 \sin i = n_2 \sin r$$
### Core Logic
{{SOL_IMG1}}
By Snell\'s law matching standard boundary normal configurations :
$$n_1 \sin(90^{\circ} - \theta_1) = n_2 \sin \theta_3 \implies n_1 \cos \theta_1 = n_2 \sin \theta_3$$ [cite: 711, 712]
Substituting the angle identity macro given [cite: 2, 713]:
$$n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin \theta_3 \implies \sin \theta_3 = \frac{1}{2} \implies \theta_3 = 30^{\circ}$$
### Step 1: Geometrical Thickness Resolution
From the block triangles geometry :
$$\tan 30^{\circ} = \frac{d/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{d}{2t}$$
$$t = \frac{d\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6\text{ cm}$$
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1373 (2024)
The distance between object and its 3 times magnified virtual image as produced by a convex lens is $20\mathrm{~cm}$. The focal length of the lens used is _______ $\mathrm{cm}$.
### Related Formula
Lens magnification formula:
$$m = \frac{v}{u}$$
Thin lens equation:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
### Core Logic
For a virtual image formed by a convex lens, both the object and the image lie on the same side of the lens. The image distance is three times the object distance:
$v = 3u$
The distance between the object and its virtual image is given as $20\mathrm{~cm}$:
$$v - u = 20\mathrm{~cm} \implies 3u - u = 20\mathrm{~cm}$$
$$2u = 20\mathrm{~cm} \implies u = 10\mathrm{~cm}$$
Applying Cartesian sign conventions: object distance $u = -10\mathrm{~cm}$ and image distance $v = -30\mathrm{~cm}$.
### Step 1: Calculate Focal Length
Substitute these values into the lens formula:
$$\frac{1}{-30} - \frac{1}{-10} = \frac{1}{f}$$
$$-\frac{1}{30} + \frac{1}{10} = \frac{1}{f}$$
$$\frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15} = \frac{1}{f} \implies f = 15\mathrm{~cm}$$
### Pattern Recognition
A virtual image from a convex lens means the object is placed inside the focal point ($u < f$). This serves as a quick sanity check for your final value ($10\mathrm{~cm} < 15\mathrm{~cm}$). ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
