Consider following statements for refraction of light through prism, when angle of deviation is minimum.
(A) The refracted ray inside prism becomes parallel to the base.
(B) Larger angle prisms provide smaller angle of minimum deviation.
(C) Angle of incidence and angle of emergence becomes equal.
(D) There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting.
(E) Angle of refraction becomes double of prism angle.
Choose the correct answer from the options given below.
A.A, C and D Only
B.B, C and D Only
C.A, B and E Only
D.B, D and E Only
Solution & Explanation
### Related Formula
For a prism of angle A$A$:
- Deviation: delta = i + e - A$\delta = i + e - A$
- Minimum deviation delta_textmin$\delta_{\text{min}}$ occurs when:
i = e quad textand quad r_1 = r_2 = fracA2$i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2}$
- Under minimum deviation, the ray inside an equilateral/isosceles prism travels symmetrically, making it parallel to the prism base.
### Core Logic
Let's check the validity of each statement:
- **Statement (A)**: The refracted ray inside the prism becomes parallel to the base at minimum deviation. (True for symmetric prisms)
- **Statement (B)**: Larger angle prisms provide smaller minimum deviation. By minimum deviation equation:
mu = fracsinleft(fracA+delta_textmin2right)sinleft(fracA2right)$\mu = \frac{\sin\left(\frac{A+\delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
As A$A$ increases, delta_textmin$\delta_{\text{min}}$ generally increases, not decreases. (False)
- **Statement (C)**: Angle of incidence i$i$ and angle of emergence e$e$ become equal (i = e$i = e$) during the minimum deviation state. (True)
- **Statement (D)**: The delta-i$\delta-i$ curve is asymmetric and parabolic-like; for any deviation delta > delta_textmin$\delta > \delta_{\text{min}}$, there are always exactly two different incident angles (i$i$ and e$e$) that yield the same deviation, except at the minimum deviation point (which has a single unique value). (True)
- **Statement (E)**: Angle of refraction r = A/2$r = A/2$, which is half of the prism angle, not double. (False)
### Step 1: Conclusion
Since statements A, C, and D are true, the correct option is (1).
### Pattern Recognition
Review the classic parabolic shape of the deviation vs. angle of incidence (delta-i$\delta-i$) graph. Notice that any horizontal line above the minimum point intersects twice (representing i$i$ and e$e$ for that deviation). Minimum deviation is the unique local minimum, where i = e$i = e$ and r_1 = r_2 = A/2$r_1 = r_2 = A/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments: Refraction through Prism
Keywords:#prism minimum deviation i=e#refracted ray parallel base#JEE Main 2025 Morning Q20#deviation angle of incidence curve
More Ray Optics and Optical Instruments Previous-Year Questions — Page 5
Q202025Total Internal Reflection
A transparent block A having refractive index mu=1.25$\mu=1.25$ is surrounded by another medium of refractive index mu=1.0$\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle theta$\theta$ as shown in figure. What is the maximum value of theta$\theta$ for which light suffers total internal reflection at the top surface of the block?
The diagram displays a light ray entering a rectangular block from a medium of lower refractive index, hitting the top wall at the critical boundary angle. [cite: 169, 170, 171]
A.tan^-1(4/3)$\tan^{-1}(4/3)$ [cite: 175]
B.tan^-1(3/4)$\tan^{-1}(3/4)$ [cite: 176]
C.sin^-1(3/4)$\sin^{-1}(3/4)$ [cite: 177]
D.cos^-1(3/4)$\cos^{-1}(3/4)$ [cite: 178]
Solution
### Related Formula
sintheta_c = fracmu_1mu_2$\sin\theta_c = \frac{\mu_1}{\mu_2}$ [cite: 810]
mu_1 sintheta = mu_2 sin r$\mu_1 \sin\theta = \mu_2 \sin r$ [cite: 807]
### Core Logic
From the boundary geometry at the top interface, the angle of refraction r$r$ at the first surface satisfies: [cite: 806]
r + theta_c = 90^circ implies r = 90^circ - theta_c$r + \theta_c = 90^{\circ} \implies r = 90^{\circ} - \theta_c$ [cite: 806]
Applying Snell's law at the first entry interface: [cite: 170, 807]
mu_1 sintheta = mu_2 sin r = mu_2 sin(90^circ - theta_c) = mu_2 costheta_c$\mu_1 \sin\theta = \mu_2 \sin r = \mu_2 \sin(90^{\circ} - \theta_c) = \mu_2 \cos\theta_c$ [cite: 172, 173, 807, 809]
Since sintheta_c = fracmu_1mu_2 = frac1.01.25 = frac45$\sin\theta_c = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{4}{5}$, we have costheta_c = sqrt1 - left(frac45right)^2 = frac35$\cos\theta_c = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$[cite: 169, 810, 811]. Substituting back into the expression: [cite: 811]
1.0 cdot sintheta = 1.25 times frac35 = frac54 times frac35 = frac34$1.0 \cdot \sin\theta = 1.25 \times \frac{3}{5} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$ [cite: 169, 811]
theta = sin^-1left(frac34
ight)$\theta = \sin^{-1}\left(\frac{3}{4}
ight)$ [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
Q142025Lenses and Magnification
A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 \, cm times 2 \, cm$2 \, cm \times 2 \, cm$ and the area of the landscape photographed is 400 \, km^2$400 \, km^2$ . The focal length of the lens in the drone camera is:
A. 1.8 cm
B. 2.8 cm
C. 2.5 cm
D. 0.9 cm
Solution
### Related Formula
Areal Magnification:
m^2 = fracA_textimageA_textobject = left(fracff+u
ight)^2 approx left(fracfu
ight)^2$m^2 = \frac{A_{\text{image}}}{A_{\text{object}}} = \left(\frac{f}{f+u}
ight)^2 \approx \left(\frac{f}{u}
ight)^2$
since object distance u = -18\ mathrmkm$u = -18\ \mathrm{km}$ is vastly larger than f$f$.
### Core Logic
Given parameters: Ray context geometry for drone camera scaling layout Q14
- Object height distance, H = 18\ mathrmkm = 18 times 10^3\ mathrmm$H = 18\ \mathrm{km} = 18 \times 10^3\ \mathrm{m}$
- Film size area, A_textimage = 2\ mathrmcm times 2\ mathrmcm = 4\ mathrmcm^2 = 4 times 10^-4\ mathrmm^2$A_{\text{image}} = 2\ \mathrm{cm} \times 2\ \mathrm{cm} = 4\ \mathrm{cm^2} = 4 \times 10^{-4}\ \mathrm{m^2}$
- Landscape area, A_textobject = 400\ mathrmkm^2 = 400 times 10^6\ mathrmm^2$A_{\text{object}} = 400\ \mathrm{km^2} = 400 \times 10^6\ \mathrm{m^2}$
Linear magnification factor:
fracyx = sqrtfracA_textimageA_textobject = sqrtfrac4 times 10^-4400 times 10^6 = sqrt10^-12 = 10^-6$\frac{y}{x} = \sqrt{\frac{A_{\text{image}}}{A_{\text{object}}}} = \sqrt{\frac{4 \times 10^{-4}}{400 \times 10^6}} = \sqrt{10^{-12}} = 10^{-6}$
Using the simple pinhole/thin lens perspective ratio:
fracfH = 10^-6 implies f = 18 times 10^3 times 10^-6 = 18 times 10^-3\ mathrmm = 1.8\ mathrmcm$\frac{f}{H} = 10^{-6} \implies f = 18 \times 10^3 \times 10^{-6} = 18 \times 10^{-3}\ \mathrm{m} = 1.8\ \mathrm{cm}$
### Pattern Recognition
For aerial satellite imaging contexts where u gg f$u \gg f$, linear sizing scales directly as fractextfilm sidetextground side = fracfH$\frac{\text{film side}}{\text{ground side}} = \frac{f}{H}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q22025Power of a Lens
What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D ? ['D' stands for dioptre]
A. 0.04
B. 0.40
C. 0.1
D. 0.01
Solution
### Related Formula
The relationship between optical power P$P$ and focal length F$F$ is given by:
F = frac1P$F = \frac{1}{P}$Relative decrease in focal length is defined as:
fracDelta FF = fracF - F'F$\frac{\Delta F}{F} = \frac{F - F'}{F}$
### Core Logic
Given initial power P = 2.5text D$P = 2.5\text{ D}$[cite: 19, 600]. After an increase of 0.1text D$0.1\text{ D}$, the new power is[cite: 19, 603]:
P' = 2.5 + 0.1 = 2.6text D$P' = 2.5 + 0.1 = 2.6\text{ D}$
### Step 1: Calculate Focal Length Change
Find the initial and final focal lengths [cite: 602, 604]:
F = frac12.5 = frac25$F = \frac{1}{2.5} = \frac{2}{5}$F' = frac12.6 = frac513$F' = \frac{1}{2.6} = \frac{5}{13}$
Now, calculate the relative decrease:
fracF - F'F = 1 - fracF'F = 1 - fracPP' = 1 - frac2.52.6 = frac0.12.6 = frac126 approx 0.04$\frac{F - F'}{F} = 1 - \frac{F'}{F} = 1 - \frac{P}{P'} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} = \frac{1}{26} \approx 0.04$
### Pattern Recognition
For a small change, we can approximate using differentiation: P = frac1F implies dP = -fracdFF^2 implies fracdFF = -fracdPP$P = \frac{1}{F} \implies dP = -\frac{dF}{F^2} \implies \frac{dF}{F} = -\frac{dP}{P}$. Thus, the relative change magnitude is frac0.12.5 = frac125 = 0.04$\frac{0.1}{2.5} = \frac{1}{25} = 0.04$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q122025Silvering of Lenses
A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-
A. 0.15 m
B. 0.10 m
C. 0.20 m
D. 0.25 m
Solution
### Related Formula
The net focal power of a silvered tracking lens system is given by:
P = 2P_L + P_M$P = 2P_{L} + P_{M}$frac1f = frac2f_L + frac1f_M$\frac{1}{f} = \frac{2}{f_{L}} + \frac{1}{f_{M}}$
### Core Logic
As shown in diagram Silvering of Lenses diagram for Q12 - JEE Main 2025 Morning, the plane flat side boundary interface has an infinite radius of curvature (R_2 = infty$R_{2} = \infty$), meaning its mirror focal component is f_M = infty implies P_M = 0$f_{M} = \infty \implies P_{M} = 0$. The power depends entirely on the refraction step:
frac1f = frac2f_L $\frac{1}{f} = \frac{2}{f_{L}} $
### Step 1: Lens Maker Formulation
Find the focal expression of the immersed lens element [cite: 91, 679]:
frac1f_L = left(fracmu_textglassmu_textliquid - 1
ight)left(frac1R
ight) = left(frac1.51.2 - 1
ight)frac1R = frac0.31.2frac1R = frac14R$\frac{1}{f_{L}} = \left(\frac{\mu_{\text{glass}}}{\mu_{\text{liquid}}} - 1
ight)\left(\frac{1}{R}
ight) = \left(\frac{1.5}{1.2} - 1
ight)\frac{1}{R} = \frac{0.3}{1.2}\frac{1}{R} = \frac{1}{4R}$
Now insert this into the total system tracking balance relation :
frac1f = 2 left(frac14R
ight) = frac12R $\frac{1}{f} = 2 \left(\frac{1}{4R}
ight) = \frac{1}{2R} $
Given the final effective concave configuration matches f = 0.2text m$f = 0.2\text{ m}$ :
frac10.2 = frac12R implies 2R = 0.2 implies R = 0.10text m$\frac{1}{0.2} = \frac{1}{2R} \implies 2R = 0.2 \implies R = 0.10\text{ m}$
### Pattern Recognition
Silvering a plano-flat back boundary means light traverses the initial curved face interface exactly twice, mapping to R = 2 cdot f cdot (mu_textrel - 1)$R = 2 \cdot f \cdot (\mu_{\text{rel}} - 1)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q152025Lens Maker's Formula
A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f_1$f_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f_2$f_{2}$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of f_1$f_{1}$ and f_2$f_{2}$ will be :-
A. 3:5
B. 1:3
C. 1:2
D. 2:3
Solution
### Related Formula
Lens Maker's Formula for a lens in a surrounding medium of refractive index mu_m$\mu_{m}$ is:
frac1f = left(fracmu_textlensmu_m - 1
ight)left(frac1R_1 - frac1R_2
ight)$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{m}} - 1
ight)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}
ight)$
### Core Logic
For a plano-convex configuration, the flat side has an infinite radius of curvature (R_2 = infty implies frac1R_2 = 0$R_{2} = \infty \implies \frac{1}{R_{2}} = 0$).
### Step 1: Evaluating Respective Focal Scales
For the first lens setup in air (mu_m = 1$\mu_{m} = 1$) :
frac1f_1 = (1.5 - 1)left(frac12 - 0
ight) = 0.5 times frac12 = frac14 implies f_1 = 4text cm$\frac{1}{f_{1}} = (1.5 - 1)\left(\frac{1}{2} - 0
ight) = 0.5 \times \frac{1}{2} = \frac{1}{4} \implies f_{1} = 4\text{ cm}$
For the second lens setup immersed inside fluid (mu_m = 1.2$\mu_{m} = 1.2$) :
frac1f_2 = left(frac1.51.2 - 1
ight)left(frac13 - 0
ight) = (1.25 - 1)frac13 = frac0.253 = frac112 implies f_2 = 12text cm$\frac{1}{f_{2}} = \left(\frac{1.5}{1.2} - 1
ight)\left(\frac{1}{3} - 0
ight) = (1.25 - 1)\frac{1}{3} = \frac{0.25}{3} = \frac{1}{12} \implies f_{2} = 12\text{ cm}$
Taking their direct ratio :
f_1 : f_2 = 4 : 12 = 1 : 3$f_{1} : f_{2} = 4 : 12 = 1 : 3$
### Pattern Recognition
Always separate the refractive index multiplier from the geometric shape factor. This lets you calculate each change independently before taking the final ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
More Ray Optics and Optical Instruments Questions — jee_main_2025_03_april_morning
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