Ray Optics and Optical Instruments Previous Year Questions — JEE Main Physics
63 past-year Ray Optics and Optical Instruments questions from JEE Main (Physics).
Q14 (2025)
Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of $12 \, \text{cm}$ , at the same distance of $18 \, \text{cm}$ from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is:
- $1 / 2$
- $2$
- $3$
- $1 / 3$
### Related Formula
1. Mirror focal length:
$$f = \frac{R}{2}$$
2. Magnification (image size relative to object size):
$$m = \frac{h_i}{h_o} = \frac{f}{f - u}$$
where $u$ is the object distance.
### Core Logic
Given parameters:
- Radius of curvature $R = 12 \ \mathrm{cm} \implies |f| = 6 \ \mathrm{cm}$
- Object distance $u = -18 \ \mathrm{cm}$
Let's calculate magnification for both mirrors:
1. **For Convex Mirror:**
- Focal length $f_{\text{convex}} = +6 \ \mathrm{cm}$ (using Cartesian sign convention)
- Magnification:
$$m_1 = \frac{f_{\text{convex}}}{f_{\text{convex}} - u} = \frac{6}{6 - (-18)} = \frac{6}{24} = \frac{1}{4}$$
Thus, image size is $\frac{1}{4} h_o$.
### Step 1: Calculate magnification of concave mirror
2. **For Concave Mirror:**
- Focal length $f_{\text{concave}} = -6 \ \mathrm{cm}$
- Magnification:
{{SOL_IMG1}}
$$m_2 = \frac{f_{\text{concave}}}{f_{\text{concave}} - u} = \frac{-6}{-6 - (-18)} = \frac{-6}{12} = -\frac{1}{2}$$
Thus, image size is $\frac{1}{2} h_o$.
### Step 2: Calculate the ratio of image sizes
Since the objects are identical (same height $h_o$), the ratio of the sizes of the images is:
{{SOL_IMG2}}
$$\text{Ratio} = \frac{|h_{i1}|}{|h_{i2}|} = \frac{|m_1|}{|m_2|} = \frac{1/4}{1/2} = \frac{1}{2}$$
Thus, the ratio of sizes is $1/2$.
### Pattern Recognition
Sees: Parallel convex vs concave mirror magnification.
Trap: Reversing the signs of focal lengths (Convex focal length is $+$, Concave is $-$ in standard coordinate systems).
Shortcut: Use direct magnification equation $m = \frac{f}{f-u}$. For convex, $m = \frac{6}{24} = \frac{1}{4}$. For concave, $m = \frac{-6}{12} = -\frac{1}{2}$. Ratio of absolute values is $\frac{1/4}{1/2} = 1/2$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q17 (2025)
A bi-convex lens has radius of curvature of both the surfaces same as $1/6$ cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( $R_1 \neq R_2$ ), without any change in lens power then possible combination of $R_1$ and $R_2$ is:
- $\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$
- $\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
- $\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
- $\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$
### Related Formula
Lens Maker's Formula:
$$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
For a bi-convex lens of equal radii ($R_1 = +R$, $R_2 = -R$):
$$\frac{1}{f} = (\mu - 1) \frac{2}{R}$$
Power $P \propto \frac{1}{f}$.
### Core Logic
For the initial lens:
- $R = \frac{1}{6} \ \mathrm{cm}$
- $\frac{1}{f_1} = (\mu - 1) \frac{2}{1/6} = 12 (\mu - 1)$
For the replacement lens ($R_1 = +R_1$, $R_2 = -R_2$):
$$\frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
Since power must be preserved ($f_1 = f_2$):
$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} = 12 \ \mathrm{cm}^{-1}$$
We need to find a combination where the sum of the reciprocals of the radii equals $12$.
### Step 1: Verify the options
Let's check each choice:
- **Option (1):** $R_1 = \frac{1}{3}$, $R_2 = \frac{1}{3}$:
$$\frac{1}{1/3} + \frac{1}{1/3} = 3 + 3 = 6 \neq 12$$
- **Option (2):** $R_1 = \frac{1}{5}$, $R_2 = \frac{1}{7}$:
$$\frac{1}{1/5} + \frac{1}{1/7} = 5 + 7 = 12 \quad \checkmark \quad \text{(Correct Combination)}$$
- **Option (3):** $R_1 = \frac{1}{3}$, $R_2 = \frac{1}{7}$:
$$\frac{1}{1/3} + \frac{1}{1/7} = 3 + 7 = 10 \neq 12$$
- **Option (4):** $R_1 = \frac{1}{6}$, $R_2 = \frac{1}{9}$:
$$\frac{1}{1/6} + \frac{1}{1/9} = 6 + 9 = 15 \neq 12$$
Thus, only Option (2) meets the physical conditions.
### Pattern Recognition
Sees: Equivalent thin lens power with modified surfaces.
Trap: Neglecting sign convention for the second spherical surface during substitution.
Shortcut: If the radii are of the form $1/n$, then the sum of $n_1 + n_2$ must equal $2 \times n_{\text{initial}}$. Since initial $n = 6$, $2 \times 6 = 12$. The only pairing whose denominators add up to 12 is $5 + 7$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q23 (2025)
A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^{circ}$ . The refractive index of the prism material is $\sqrt{2}$ . The angle of incidence (in degrees) is ______.
### Related Formula
1. Prism Formula relating refractive index to minimum deviation:
$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
2. Under the condition of minimum deviation:
$$i = \frac{A + \delta_m}{2}$$
Thus, the formula simplifies to:
$$\mu = \frac{\sin i}{\sin(A/2)}$$
### Core Logic
Given parameters:
- Angle of prism $A = 60^\circ$
- Refractive index of material $\mu = \sqrt{2}$
Substitute the parameters into the simplified formula:
$$\sqrt{2} = \frac{\sin i}{\sin(60^\circ / 2)}$$
$$\sqrt{2} = \frac{\sin i}{\sin(30^\circ)}$$
### Step 1: Solve for angle of incidence
Since $\sin(30^\circ) = 0.5$:
$$\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$$
Solving for $i$:
$i = 45^\circ$
Thus, the angle of incidence is $45$ degrees.
### Pattern Recognition
Sees: Prism minimum deviation and refracting index relations.
Trap: Mistaking the computed angle of incidence ($i$) for the angle of minimum deviation ($\delta_m$).
Shortcut: Snell's law at the symmetrical boundary reduces to $\mu = \sin i / \sin(A/2)$. Because $A = 60^\circ$, the denominator is $\sin 30^\circ = 1/2$. Therefore, $\sin i = 0.5 \times \sqrt{2} = 1/\sqrt{2} \implies i = 45^\circ$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q79 (2025)
A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:
{{IMG1}}
- $-\frac{\alpha}{2}$
- $-45^{\circ}$
- $+45^{\circ}$
- $-\alpha$
### Related Formula
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$m = \frac{v}{u}$$
$$m_L = \frac{dv}{du} = m^2$$
### Core Logic
Let the pole of the lens be the origin. Point $A$ of the slanted object lies on the principal axis at $u = -30\mathrm{~cm}$ from the convex lens ($f = +20\mathrm{~cm}$).
Let's locate the image of $A$:
$$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \implies v = +60\mathrm{~cm}$$
Thus, the transverse magnification $m$ at point $A$ is:
$$m = \frac{v}{u} = \frac{60}{-30} = -2$$
Since the longitudinal extension of the object is small ($du = 1\mathrm{~cm}$ along the axis):
$$dv = m^2 du = (-2)^2 \times 1 = 4\mathrm{~cm}$$
The height of the object at point $B$ is $h_o = 2\mathrm{~cm}$. Its image height is:
$$h_i = m \cdot h_o = (-2) \times 2 = -4\mathrm{~cm}$$
Now, compute the angle $\beta$ made by the image with the principal axis:
$$\tan\beta = \frac{h_i}{dv} = \frac{-4\mathrm{~cm}}{4\mathrm{~cm}} = -1$$
$$\beta = -45^{\circ}$$
### Step 1: Final Conclusion
The angle made by the image with the principal axis is $-45^{\circ}$.
### Pattern Recognition
For small objects tilted with respect to the principal axis:
1. Axial displacement scales by $m^2$.
2. Transverse height scales by $m$.
3. Slope scales by $\frac{m}{m^2} = \frac{1}{m}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics
Q17 (2025)
A spherical surface separates two media of refractive indices 1 and 1.5 as shown in the figure. Distance of the image of an object 'O', is:
(C is the center of curvature of the spherical surface and R is the radius of curvature)
{{IMG1}}
- $0.24\mathrm{~m}$ right to the spherical surface
- $0.4\mathrm{~m}$ left to the spherical surface
- $0.24\mathrm{~m}$ left to the spherical surface
- $0.4\mathrm{~m}$ right to the spherical surface
### Related Formula
$$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$$
### Core Logic
From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin:
- Refractive index of first medium, $\mu_1 = 1.0$
- Refractive index of second medium, $\mu_2 = 1.5$
- Object distance, $u = -0.2\mathrm{~m}$ (left of surface)
- Radius of curvature, $R = +0.4\mathrm{~m}$ (convex surface towards first medium, center $C$ lies in second medium)
Applying the formula for refraction at a spherical interface:
$$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4}$$
$$\frac{1.5}{v} + 5.0 = \frac{0.5}{0.4} = 1.25$$
$$\frac{1.5}{v} = 1.25 - 5.0 = -3.75$$
$$v = \frac{1.5}{-3.75} = -0.4\mathrm{~m}$$
The negative sign indicates that the image is formed to the left of the spherical refracting surface.
### Step 1: Final Conclusion
The image of object 'O' is formed $0.4\mathrm{~m}$ left to the spherical surface.
### Pattern Recognition
Ensure to strictly implement the coordinate sign conventions: the direction of incident light is positive. Since light goes from left to right, left-side points have a negative coordinate, and right-side points have a positive coordinate.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics
