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Two light beams fall on a transparent material block at point 1 and 2 with angle theta_1 and 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 and theta_1 = theta_2 = cos^-1left(fracn_22n_1right) , where refractive index of the block n_2 > refractive index of the outside medium n_1 , then the thickness of the block is ______ cm. [cite: 1, 2]
Refraction diagram for Q23 - JEE Main 2025 Morning
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.

Numerical Answer Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

### Related Formula n_1 sin i = n_2 sin r ### Core Logic
Refraction explanation geometric mapping
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 [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 ### Step 1: Geometrical Thickness Resolution From the block triangles geometry : tan 30^circ = fracd/2t implies frac1sqrt3 = fracd2t t = fracdsqrt32 = frac4sqrt3 cdot sqrt32 = 6text cm ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 3

Q13 2025 Lens Maker's Formula
A convex lens made of glass (refractive index = 1.5 ) has focal length 24 \, textcm in air. When it is totally immersed in water (refractive index = 1.33 ), its focal length changes to:
  • A. 72mathrm~cm
  • B. 96mathrm~cm
  • C. 24mathrm~cm
  • D. 48mathrm~cm

Solution

### Related Formula frac1f = left(fracmu_gmu_m - 1right)left(frac1R_1 - frac1R_2right) ### Core Logic In air (mu_m = 1): frac124 = (1.5 - 1) cdot K = 0.5 K implies K = frac112 quad dots (i) In water (mu_m = 1.33 = frac43): frac1f' = left(frac1.54/3 - 1right) cdot K = left(frac4.54 - 1right) cdot K = frac18 K quad dots (ii) Dividing equation (i) by equation (ii): fracf'24 = frac0.51/8 = 4 f' = 24 times 4 = 96mathrm~cm ### Pattern Recognition Standard relation for standard glass lens (mu=1.5) immersed in water (mu=4/3): the focal length always becomes exactly 4 times its original value in air (f_textwater = 4 f_textair). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments
Q5 2025 Total Internal Reflection
A hemispherical vessel is completely filled with a liquid of refractive index mu . A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is
Total Internal Reflection diagram for Q5 - JEE Main 2025 Morning
A coin at the base of a hemispherical liquid filled container viewed from the grazing edge point E.
  • A. sqrt3
  • B. frac32
  • C. sqrt2
  • D. fracsqrt32

Solution

### Related Formula sin mathrmc = frac1mu ### Core Logic To see the coin from edge point mathrmE at grazing emergency, the light ray travelling from mathrmO to mathrmE must strike the flat upper boundary at the critical angle.
Ray tracing verification layout for Q5
A coin at the base of a hemispherical liquid filled container viewed from the grazing edge point E.
Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: theta = mathrmc = 45^circ Substituting this value into the critical value expression: mu = frac1sin 45^circ = sqrt2 ### Step 1: Final Conclusion The minimum refractive index required is sqrt2, matching option (3). ### Pattern Recognition Grazing boundary ray vision configurations dictate evaluating the specific systemic geometric configuration to compute the critical boundary angle. Here, the radius profile fixes theta = 45^circ deterministically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments
Q20 2025 Prism and Dispersion
A thin prism mathrmP_1 with angle 4^circ made of glass having refractive index 1.54, is combined with another thin prism mathrmP_2 made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism mathrmP_2 in degrees is
  • A. 4
  • B. 3
  • C. 16/3
  • D. 1.5

Solution

### Related Formula delta = (mu - 1)mathrmA ### Core Logic To achieve dispersion without deviation, the net deviation produced by the prism combination must be zero: delta_textnet = 0 implies (mu_1 - 1)mathrmA_1 - (mu_2 - 1)mathrmA_2 = 0 Substituting the given parameters into the equation: (1.54 - 1) cdot 4^circ - (1.72 - 1)mathrmA_2 = 0 0.54 cdot 4 = 0.72 cdot mathrmA_2 mathrmA_2 = frac2.160.72 = 3^circ ### Step 1: Final Angle Value The required angle for the second thin prism is 3^circ, which matches option (2). ### Pattern Recognition For zero deviation conditions using thin components, balance the deviation equations directly: (mu-1)mathrmA = (mu'-1)mathrmA'. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments
Q18 2025 Lens Maker's Formula
The radii of curvature for a thin convex lens are 10mathrm~cm and 15mathrm~cm respectively. The focal length of the lens is 12mathrm~cm. The refractive index of the lens material is:
  • A. 1.2
  • B. 1.4
  • C. 1.5
  • D. 1.8

Solution

### Related Formula Lens Maker's Formula: frac1f = (mu - 1) left(frac1R_1 - frac1R_2right) where, f = focal length of the lens, mu = refractive index of the material, R_1, R_2 = radii of curvature with standard Cartesian sign convention. ### Core Logic For a thin bi-convex lens, using standard coordinate conventions: - R_1 = +10mathrm~cm (positive since first surface centers to the right of light trajectory), - R_2 = -15mathrm~cm (negative since second surface centers to the left), - Focal length f = +12mathrm~cm. ### Step 1: Substituting in the Equation Substitute the values into Lens Maker's formula: frac112 = (mu - 1) left(frac110 - frac1-15right) frac112 = (mu - 1) left(frac110 + frac115right) frac112 = (mu - 1) left(frac3 + 230right) frac112 = (mu - 1) left(frac530right) = (mu - 1) left(frac16right) mu - 1 = frac612 = 0.5 implies mu = 1.5 ### Pattern Recognition Convex lenses always have opposite signs for R_1 and R_2. The term left(frac1R_1 - frac1R_2right) is additive: left(frac1|R_1| + frac1|R_2|right). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments
Q20 2025 Minimum Deviation in Prism
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

### Related Formula For a prism of angle A: - Deviation: delta = i + e - A - Minimum deviation delta_textmin occurs when: i = e quad textand quad r_1 = r_2 = fracA2 - 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) As A increases, delta_textmin generally increases, not decreases. (False) - **Statement (C)**: Angle of incidence i and angle of emergence e become equal (i = e) during the minimum deviation state. (True) - **Statement (D)**: The delta-i curve is asymmetric and parabolic-like; for any deviation delta > delta_textmin, there are always exactly two different incident angles (i and 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, 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) graph. Notice that any horizontal line above the minimum point intersects twice (representing i and e for that deviation). Minimum deviation is the unique local minimum, where i = e and r_1 = r_2 = A/2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments: Refraction through Prism

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