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 2.
Q171 (2025)
Light from a point source in air falls on a spherical glass surface (refractive index, $\mu=1.5$ and radius of curvature $=50$ cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ________ m.
### Related Formula
The refraction equation at a single spherical interface is given by:
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
where:
- $\mu_1$ is the refractive index of the initial medium (air, $\mu_1 = 1.0$)
- $\mu_2$ is the refractive index of the second medium (glass, $\mu_2 = 1.5$)
- $u$ is the object distance
- $v$ is the image distance
- $R$ is the radius of curvature
### Core Logic
Given parameters:
- $\mu_1 = 1.0$, $\mu_2 = 1.5$
- Radius of curvature $R = +50\mathrm{~cm}$
- Image distance $v = +200\mathrm{~cm}$ (real image inside glass)
{{SOL_IMG1}}
### Step 1: Substitute parameters into refraction formula
$$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1.0}{50}$$
$$\frac{3}{400} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}$$$
### Step 2: Solve for object distance ($u$)
$$-\frac{1}{u} = \frac{1}{100} - \frac{3}{400}$$
$$-\frac{1}{u} = \frac{4 - 3}{400} = \frac{1}{400}$$
$$u = -400\mathrm{~cm} = -4\mathrm{~m}$$
The magnitude of the distance of the light source is $4\mathrm{~m}$.
### Pattern Recognition
Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $v$ and $R$ are positive, whereas $u$ is negative.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q9 (2025)
A monochromatic light of frequency $5\times10^{14}\mathrm{~Hz}$ travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :
- 300 nm
- 600 nm
- 400 nm
- 500 nm
### Related Formula
For light propagation, wave velocity, frequency, and wavelength are related by:
$$v = f \lambda \Rightarrow \lambda_{\text{air}} = \frac{c}{f}$$
When light passes into a medium of refractive index $\mu$, the frequency remains constant, but the wavelength scales down to:
$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$$
### Core Logic
Given parameters:
- Frequency $f = 5 \times 10^{14}\mathrm{~Hz}$
- Speed of light in vacuum/air $c \approx 3 \times 10^8\mathrm{~m/s}$
- Refractive index of medium $\mu = 2$
### Step 1: Calculate Wavelength in Air (Vacuum)
$$\lambda_{\text{air}} = \frac{3 \times 10^8\mathrm{~m/s}}{5 \times 10^{14}\mathrm{~Hz}} = 0.6 \times 10^{-6}\mathrm{~m} = 600\mathrm{~nm}$$
### Step 2: Calculate Refracted Wavelength in Medium
$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu} = \frac{600\mathrm{~nm}}{2} = 300\mathrm{~nm}$$
### Pattern Recognition
Remember: Frequency is a source characteristic and never changes during refraction. Speed and wavelength both decrease by a factor of $\mu$ inside the medium.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q248 (2025)
A container contains a liquid with refractive index of 1.2 up to a height of $60~\mathrm{cm}$ and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is $40~\mathrm{cm}$ . The value of H is ______$\mathrm{cm}$ . (Consider liquids are immisible)
### Related Formula
The apparent shift $\Delta x$ in depth through multiple immiscible liquid layers viewed normally is the sum of the individual layer shifts:
$$\Delta x = \sum d_i \left( 1 - \frac{1}{\mu_i} \right)$$
{{SOL_IMG1}}
### Core Logic
For two liquid layers:
- Layer 1: $d_1 = 60 \mathrm{~cm}$, $\mu_1 = 1.2$
- Layer 2: $d_2 = H \mathrm{~cm}$, $\mu_2 = 1.6$
Total apparent shift is given as $\Delta x = 40 \mathrm{~cm}$.
### Step 1: Set Up and Solve the Equation
Substitute the parameters into the equation:
$$40 = 60 \left( 1 - \frac{1}{1.2} \right) + H \left( 1 - \frac{1}{1.6} \right)$$
Calculate the fractional factors:
$$1 - \frac{1}{1.2} = 1 - \frac{5}{6} = \frac{1}{6}$$
$$1 - \frac{1}{1.6} = 1 - \frac{5}{8} = \frac{3}{8}$$
Substitute back:
$$40 = 60 \left( \frac{1}{6} \right) + H \left( \frac{3}{8} \right)$$
$$40 = 10 + \frac{3}{8} H \implies \frac{3}{8} H = 30$$
$$H = \frac{30 \times 8}{3} = 80 \mathrm{~cm}$$$
### Pattern Recognition
Sees: Two immiscible liquid layers with normal viewing shift.
Shortcut: First layer has real depth 60, index 1.2 $\implies$ apparent shift is $60 \times (1 - 5/6) = 10 \mathrm{~cm}$. Since total shift is 40, the second layer must contribute $30 \mathrm{~cm}$ of shift. Thus, $H \times (1 - 5/8) = 30 \implies H \times (3/8) = 30 \implies H = 80 \mathrm{~cm}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q8 (2025)
A lens having refractive index 1.6 has focal length of $12\mathrm{cm}$ , when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)
- $355\mathrm{mm}$
- $288\mathrm{mm}$
- $555\mathrm{mm}$
- $655\mathrm{mm}$
### Related Formula
Lens Maker's Formula in a surrounding medium with refractive index $\mu_m$ is:
$$\frac{1}{f} = \left( \frac{\mu_L}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
### Core Logic
In air ($\mu_m = 1$):
$$\frac{1}{12} = (1.6 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
$$\frac{1}{12} = 0.6 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{12 \times 0.6} = \frac{10}{72}$$
### Step 1: Calculate Focal Length in Water
In water ($\mu_m = 1.28$):
$$\frac{1}{f_w} = \left( \frac{1.6}{1.28} - 1 \right) \left( \frac{10}{72} \right)$$
Simplify the relative index factor:
$$\frac{1.6}{1.28} = \frac{160}{128} = 1.25$$
$$\frac{1}{f_w} = (1.25 - 1) \left( \frac{10}{72} \right) = 0.25 \times \frac{10}{72} = \frac{1}{4} \times \frac{10}{72} = \frac{10}{288}$$
$$f_w = 28.8 \mathrm{~cm} = 288 \mathrm{~mm}$$$
### Pattern Recognition
Sees: Lens index $\mu_L = 1.6$, focal length in air $f_a$, and focal length in medium $f_m$.
Shortcut: Use the ratio of focal lengths directly:
$$\frac{f_m}{f_a} = \frac{\mu_L - 1}{\frac{\mu_L}{\mu_m} - 1} = \frac{0.6}{\frac{1.6}{1.28} - 1} = \frac{0.6}{0.25} = 2.4$$
$$f_m = 2.4 \times 12 = 28.8 \mathrm{~cm} = 288 \mathrm{~mm}$$
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q10 (2025)
Two thin convex lenses of focal lengths $30~\mathrm{cm}$ and $10~\mathrm{cm}$ are placed coaxially, $10~\mathrm{cm}$ apart. The power of this combination is :
- $5 \mathrm{D}$
- $1 \mathrm{~D}$
- $20\mathrm{D}$
- $10\mathrm{D}$
### Related Formula
The equivalent focal length $f_{\text{eq}}$ of two thin coaxially aligned lenses separated by distance $d$ is given by:
$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$
The equivalent power in diopters ($D$) when focal lengths are in meters is:
$$P = \frac{1}{f_{\text{eq}}}$$
### Core Logic
Given parameters:
- $f_1 = 30 \mathrm{~cm} = 0.3 \mathrm{~m}$
- $f_2 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
- $d = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
### Step 1: Calculate Power
Substitute parameters into the equivalent focal length equation:
$$\frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{0.3 \times 0.1}$$
$$\frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + 10 - \frac{1}{0.3}$$
$$\frac{1}{f_{\text{eq}}} = 10 \mathrm{~m}^{-1}$$
$$P = 10 \mathrm{~D}$$
### Pattern Recognition
Sees: Lenses separated by distance $d$ where $d = f_2$.
Shortcut: Notice that $d = f_2 = 10 \mathrm{~cm}$. When the separation distance between two thin lenses equals the focal length of the second lens, the equivalent power simplifies directly to $P = P_2 = 1/f_2 = 10 \mathrm{~D}$ since the terms $1/f_1$ and $d/(f_1 f_2)$ cancel out.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
