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 6
Q82025Total Internal Reflection
At the interface between two materials having refractive indices n_1$n_1$ and n_2$n_2$ , the critical angle for reflection of an em wave is theta_1C$\theta_{1C}$ . The n_2$n_2$ material is replaced by another material having refractive index n_3$n_3$ , such that the critical angle at the interface between n_1$n_1$ and n_3$n_3$ materials is theta_2C$\theta_{2C}$ . If n_3 > n_2 > n_1$n_3 > n_2 > n_1$ ; fracn_2n_3 = frac25$\frac{n_2}{n_3} = \frac{2}{5}$ and sin theta_2C - sin theta_1C = frac12$\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$ , then theta_1C$\theta_{1C}$ is [cite: 1, 2]
### Core Logic
Critical angle setups dictate [cite: 613, 614]:
sin theta_1C = fracn_1n_2, quad sin theta_2C = fracn_1n_3$\sin \theta_{1C} = \frac{n_1}{n_2}, \quad \sin \theta_{2C} = \frac{n_1}{n_3}$ [cite: 613, 614]
Given sin theta_2C - sin theta_1C = frac12$\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$ [cite: 2, 615]:
fracn_1n_3 - fracn_1n_2 = frac12 implies n_1 left(frac25 - 1right) = fracn_22$\frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2} \implies n_1 \left(\frac{2}{5} - 1\right) = \frac{n_2}{2}$ [cite: 617, 619, 621]
fracn_1n_2 = -frac56$\frac{n_1}{n_2} = -\frac{5}{6}$
This produces an impossible negative value for a refractive index ratio, meaning the data contains a contradiction. This question was dropped by NTA.
Q172025Lenses
Let mathbfu$\mathbf{u}$ and mathbfv$\mathbf{v}$ be the distances of the object and the image from a lens of focal length f$f$ . The correct graphical representation of mathbfu$\mathbf{u}$ and mathbfv$\mathbf{v}$ for a convex lens when |mathbfu| > f$|\mathbf{u}| > f$ , is [cite: 1, 2]
A. {{IMG_OPT1}}
B. {{IMG_OPT2}}
C. {{IMG_OPT3}}
D. {{IMG_OPT4}}
Solution
### Related Formula
(u+f)(v-f) = f^2$(u+f)(v-f) = f^2$
### Core Logic
Lens formula explanation curves
By rewriting the lens equation coordinates with respect to the focal focuses, we track a rectangular hyperbola structure. In the quadrant satisfying |u| > f$|u| > f$, the matching curvature coordinates belong strictly to the profile shown in option (2).
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q232025Refraction
Two light beams fall on a transparent material block at point 1 and 2 with angle theta_1$\theta_{1}$ and theta_2$\theta_{2}$ , 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 = 4sqrt3 mathrm~cm$d = 4\sqrt{3} \mathrm{~cm}$ and theta_1 = theta_2 = cos^-1left(fracn_22n_1right)$\theta_{1} = \theta_{2} = \cos^{-1}\left(\frac{n_{2}}{2n_{1}}\right)$ , where refractive index of the block n_2 >$n_{2} >$ refractive index of the outside medium n_1$n_{1}$ , then the thickness of the block is ______ cm. [cite: 1, 2]
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.
Numerical Answer.Answer: 6 to 6
Solution
### Related Formula
n_1 sin i = n_2 sin r$n_1 \sin i = n_2 \sin r$
### Core Logic
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.
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$n_1 \sin(90^{\circ} - \theta_1) = n_2 \sin \theta_3 \implies n_1 \cos \theta_1 = n_2 \sin \theta_3$ [cite: 711, 712]
Substituting the angle identity macro given [cite: 2, 713]:
n_1 left(fracn_22n_1right) = n_2 sin theta_3 implies sin theta_3 = frac12 implies theta_3 = 30^circ$n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin \theta_3 \implies \sin \theta_3 = \frac{1}{2} \implies \theta_3 = 30^{\circ}$
### Step 1: Geometrical Thickness Resolution
From the block triangles geometry :
tan 30^circ = fracd/2t implies frac1sqrt3 = fracd2t$\tan 30^{\circ} = \frac{d/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{d}{2t}$t = fracdsqrt32 = frac4sqrt3 cdot sqrt32 = 6text cm$t = \frac{d\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6\text{ cm}$
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
More Ray Optics and Optical Instruments Questions — jee_main_2025_03_april_morning
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