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 8.
Q38 (2024)
If the distance between object and its two times magnified virtual image produced by a curved mirror is $15\text{ cm}$, the focal length of the mirror must be:
- $15\text{ cm}$
- $-12\text{ cm}$
- $-10\text{ cm}$
- $10/3\text{ cm}$
### Related Formula
Magnification formula for spherical mirrors:
$$m = -\frac{v}{u}$$
Mirror formula:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
where:
* $u$ is the object distance
* $v$ is the image distance
* $f$ is the focal length
### Core Logic
Since the image is magnified ($m = 2$) and virtual, the mirror must be concave ($f < 0$).
Let us use standard coordinate geometry signs: object is on the left ($u$ is negative, say $-u_0$), virtual image is on the right ($v$ is positive, say $+v_0$).
Given magnification:
$$m = 2 = -\frac{v}{u} \implies v = -2u$$
In terms of magnitudes:
$v_0 = 2u_0$
### Step 1: Use Distance Condition
The distance between the object and virtual image is $15\text{ cm}$. Since the object is in front of the mirror and the virtual image is behind it:
$$\text{Distance} = u_0 + v_0 = 15\text{ cm}$$
Substitute $v_0 = 2u_0$:
$$u_0 + 2u_0 = 15 \implies 3u_0 = 15 \implies u_0 = 5\text{ cm}$$
Thus:
* $u_0 = 5\text{ cm} \implies u = -5\text{ cm}$
* $v_0 = 10\text{ cm} \implies v = +10\text{ cm}$
{{SOL_IMG1}}
### Step 2: Calculate Focal Length
Using the mirror formula:
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
$$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$$
$$\frac{1}{f} = \frac{1 - 2}{10} = -\frac{1}{10} \implies f = -10\text{ cm}$$
Thus, the focal length is $-10\text{ cm}$.
### Pattern Recognition
Virtual and magnified image → always concave mirror. Distance $D$ between object and virtual image is given by $D = |u| + v$. Since $v = m|u|$, we can simplify to $|u| = \frac{D}{m+1}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1554 (2024)
Two immiscible liquids of refractive indices $\frac{3}{2}$ and $\frac{8}{5}$ respectively are put in a beaker. The height of each column is $6\text{ cm}$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4}\text{ cm}$. The value of $\alpha$ is ______.
### Related Formula
$$h_{\text{apparent}} = \sum \frac{h_i}{\mu_i}$$
### Core Logic
Sum the contributions of both shifting mediums:
$$h_{\text{apparent}} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$$
Given $h_1 = h_2 = 6\text{ cm}$, $\mu_1 = \frac{3}{2}$, $\mu_2 = \frac{8}{5}$:
### Step 1: Compute fraction value
$$h_{\text{apparent}} = \frac{6}{3/2} + \frac{6}{8/5} = 4 + \frac{30}{8} = 4 + \frac{15}{4}$$
$$h_{\text{apparent}} = \frac{16 + 15}{4} = \frac{31}{4}\text{ cm}$$
### Step 2: Match to target format
Comparing with $\frac{\alpha}{4}$ directly yields:
$\alpha = 31$
### Pattern Recognition
Apparent depth across multi-layered planar mediums expands additively via separate individual medium thickness-to-refractive-index ratios.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q33 (2024)
If the refractive index of the material of a prism is $\cot\left(\frac{A}{2}\right)$, where $A$ is the angle of prism, then the angle of minimum deviation will be:
- $\pi - 2A$
- $\frac{pi}{2} - 2A$
- $\pi - A$
- $\frac{pi}{2} - A$
### Related Formula
$$\mu = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
### Core Logic
Given $\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Equating this to the prism formula:
$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
$$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$
### Step 1: Trigonometric Substitution
Convert the cosine term to sine:
$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$
$$\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\text{min}}}{2}$$
$$\pi - A = A + \delta_{\text{min}}$$
$$\delta_{\text{min}} = \pi - 2A$$
### Pattern Recognition
When $\mu$ equals a cotangent function of half-angle, the sine dynamic simplifications lead directly to linear functions involving complements of angle $A$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q1624 (2024)
A convex mirror of radius of curvature $30 \mathrm{~cm}$ forms an image that is half the size of the object. The object distance is:
- $-15 \mathrm{~cm}$
- $45 \mathrm{~cm}$
- $-45 \mathrm{~cm}$
- $15 \mathrm{~cm}$
### Related Formula
The magnification ($m$) of a spherical mirror is given by:
$$m = \frac{f}{f - u}$$
where,
$f$ = focal length ($R/2$)
$u$ = object distance
### Core Logic
Given radius of curvature, $R = 30 \mathrm{~cm}$.
For a convex mirror:
$$f = +\frac{R}{2} = +15 \mathrm{~cm}$$
Since a convex mirror always forms a virtual, erect, and diminished image of a real object, the magnification $m$ must be positive:
$$m = +\frac{1}{2}$$
{{SOL_IMG1}}
### Step 1: Solve for Object Distance
Using the magnification formula:
$$+\frac{1}{2} = \frac{15}{15 - u}$$
$$15 - u = 30 \implies u = -15 \mathrm{~cm}$$
Thus, the object distance is $-15 \mathrm{~cm}$.
### Pattern Recognition
Remember: Convex mirrors produce *only* virtual images for real objects, which means $m$ is always positive and less than $1$. If the question mentioned a concave mirror with a diminished image of half size, the image would be real, and $m$ would be negative.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q45 (2024)
A biconvex lens of refractive index 1.5 has a focal length of $20 \mathrm{~cm}$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
- $-16 \mathrm{~cm}$
- $-160 \mathrm{~cm}$
- $+160 \mathrm{~cm}$
- $+16 \mathrm{~cm}$
### Related Formula
From the Lens Maker's Formula:
$$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
Taking the ratio of focal length in liquid medium ($f_m$) to focal length in air ($f_a$):
$$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{\mu_1 - \mu_m}$$
where,
$\mu_1$ = refractive index of the lens material = 1.5
$\mu_m$ = refractive index of the liquid medium = 1.6
$f_a$ = focal length in air = $20 \mathrm{~cm}$
### Core Logic
Substitute the parameters into the relative ratio template equation:
$$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$$
### Step 1: Simplify and Compute
$$$\frac{f_m}{20} = \frac{0.5 \times 1.6}{-0.1}$$
$$\frac{f_m}{20} = \frac{0.8}{-0.1} = -8$$
$$f_m = -8 \times 20 = -160 \mathrm{~cm}$$
Therefore, the focal length in the liquid is $-160 \mathrm{~cm}$.
### Pattern Recognition
Notice that since the surrounding liquid medium has a higher refractive index than the lens material itself ($\mu_m \gt \mu_1$), the sign of the focal length flips from positive to negative. The convex lens behaves as a diverging lens inside this specific liquid.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
