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

To see the coin from edge point E at grazing emergency, the light ray travelling from mathrmO to mathrmE must strike the flat upper boundary at the critical angle. Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: Substituting this value into the critical value expression: The minimum refractive index required is sqrt(2), matching option (3). 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 deterministically. Class 12 Physics: Ray Optics and Optical Instruments

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 in

To see the coin from edge point E at grazing emergency, the light ray travelling from mathrmO to mathrmE must strike the flat upper boundary at the critical angle. Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: Substituting this value into the critical value expression: The minimum refractive index required is sqrt(2), matching option (3). Grazing boundary ray vision configurations dictate evaluating the specific systemic geome

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

To see the coin from edge point E at grazing emergency, the light ray travelling from mathrmO to mathrmE must strike the flat upper boundary at the critical angle. Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: Substituting this value into the critical value expression: The minimum refractive index required is sqrt(2), matching option (3). 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 deterministically. Class 12 Physics: Ray Optics and Optical Instruments

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 in

To see the coin from edge point E at grazing emergency, the light ray travelling from mathrmO to mathrmE must strike the flat upper boundary at the critical angle. Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: Substituting this value into the critical value expression: The minimum refractive index required is sqrt(2), matching option (3). Grazing boundary ray vision configurations dictate evaluating the specific systemic geome

A hemispherical vessel is completely filled with a liquid of refractive index mu . A small coin is kept at the lowest…

Year	2026	2025	2024	Total
Questions	19	34	10	63

Q48 jee_main_2026_22_january_morning Refraction at Spherical Surfaces

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is \_\_\_\_ cm.

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

(μ₂)/(v) - (μ₁)/(u) = (μ₂ - μ₁)/(R)

Core Logic

Solution spherical refraction diagram for Q48 - JEE Main 2026 Morning

For parallel beam incident from air ( $u = \infty$ $u = \infty$):

(1.5)/(v) - (1)/(\infty) = (1.5 - 1.0)/(50) v = 150 cm

Distance from center of curvature:

x = v - R = 150 - 50 = 100 cm

Pattern Recognition

Sees: Refraction at convex spherical surface with parallel incident beam. Shortcut: Apply spherical surface refraction formula with $u=\infty$ $u=\infty$. Check: Numerical answer is 100 cm. ✓

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q31 jee_main_2026_22_january_evening Concave Mirror Experiments

In parallax method for the determination of focal length of a concave mirror, the object should always be placed :

A. between the focus (F) and the centre of curvature (C) of the mirror ONLY
B. at any point beyond the focus (F) of the mirror
C. beyond the centre of the curvature (C) of the mirror ONLY
D. between the pole (P) and the focus (F) of the concave mirror ONLY

Solution

Related Formula

(1)/(f) = (1)/(v) + (1)/(u)

Core Logic

To locate an image using the optical bench parallax method, a real image must be formed in space in front of the mirror so that an optical pin can be aligned with it.

For a concave mirror, real images are formed only when the object is located beyond the principal focus ( $u > f$ $u > f$). If placed between pole and focus, a virtual image is formed behind the mirror, preventing parallax removal using standard optical pins.

Step 1: Final Conclusion

Hence, the object can be placed at any point beyond the focus (F) of the mirror.

Pattern Recognition

Experimental optics rule: Parallax method requires real image formation $\implies$ object must be placed beyond focus ( $u > f$ $u > f$).

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q26 jee_main_2026_23_january_morning Dispersion by a Prism

A thin prism with angle

5°

of refractive index 1.72 is combined with another prism of refractive index 1.9 to produce dispersion without deviation. The angle of second prism is

A. $4.5°$ $4.5^{\circ}$
B. $6°$ $6^{\circ}$
C. $4°$ $4^{\circ}$
D. $5°$ $5^{\circ}$

Solution

Related Formula

δₙₑₜ = 0

δ₁ + δ₂ = 0

(μ₁ - 1)A₁ + (μ₂ - 1)A₂ = 0

Core Logic

For a combination of two prisms producing dispersion without deviation, the net angular deviation must be zero. The opposing orientations mean their deviations will cancel out algebraically.

Step 1: Final Calculation

A₂ = (μ₁ - 1)A₁(μ₂ - 1)

A₂ = ((1.72 - 1))/((1.9 - 1)) \times 5° = (0.72)/(0.9) \times 5° = 4°

Pattern Recognition

Sees: "dispersion without deviation" → net deviation is zero. Shortcut: $(μ₁ - 1)A₁ = (μ₂ - 1)A₂$ . Inverse relation between angle and refractive index difference.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q36 jee_main_2026_23_january_morning Total Internal Reflection

Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are

2.4 \times 10⁸ m/s

and

2.7 \times 10⁸ m/s

respectively, then the value of critical angle is:

A. $⁻¹( 3\sqrt(13))$
B. $⁻¹((9)/(8))$ $\sin^{-1}\left(\frac{9}{8}\right)$
C. $⁻¹( 8\sqrt(17))$
D. $⁻¹((8)/(9))$ $\cos^{-1}\left(\frac{8}{9}\right)$

Solution

Related Formula

μA c = μB 90°

μ = (c)/(v)

Core Logic

For total internal reflection at the critical angle, the angle of refraction is $90°$ $90^{\circ}$. Snell's law gives $c = μB/μA$ $\sin c = \mu_{B}/\mu_{A}$. Since refractive index is inversely proportional to velocity, this ratio is equal to the ratio of velocities $vA/vB$ $v_{A}/v_{B}$.

Step 1: Velocity Ratio to Sine

c = vAvB

c = 2.4 \times 10⁸2.7 \times 10⁸ = (24)/(27) = (8)/(9)

Step 2: Triangle Mapping to Tangent

We need the answer in terms of $⁻¹$ $\tan^{-1}$ based on the options. From a right triangle with Opposite = 8, Hypotenuse = 9:

Adjacent = 9² - 8² = \sqrt(81 - 64) = \sqrt(17)

c = OppositeAdjacent = 8\sqrt(17)

c = ⁻¹( 8\sqrt(17))

Pattern Recognition

Sees: "total internal reflection" + "speeds given" → skip calculating refractive indices, directly jump to $c = vdenser / vᵣₐᵣₑᵣ$ . Use Pythagorean identity if options differ in trigonometric function.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q40 jee_main_2026_23_january_evening Prism and Total Internal Reflection

A prism of angle

75°

and refractive index

\sqrt(3)

is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be ____. (

15° = 0.25

and

25° = 0.43

)

A. $between 15° and 20°$
B. $15°$ $15^{\circ}$
C. $> 25°$ $> 25^{\circ}$
D. $< 15°$ $< 15^{\circ}$

Solution

Related Formula

$r₁ + r₂ = A$ $r_1 + r_2 = A$ Snell's Law: $μ₁ i = μ₂ r$ $\mu_1 \sin i = \mu_2 \sin r$ Condition for TIR: $θ > θC$ $\theta > \theta_C$, where $θC = μᵣₐᵣₑᵣμdenser$

Core Logic

Prism and Total Internal Reflection diagram for Q40 - JEE Main 2026 Evening

Given prism angle $A = 75°$ $A = 75^{\circ}$. For TIR at the back surface, light travels from prism ( $μ = \sqrt(3)$ $\mu = \sqrt{3}$) to film ( $μ = 1.5$ $\mu = 1.5$).

r₁ + r₂ = 75°

Step 1: Critical Angle for TIR

Applying Snell's Law for critical angle at the back surface:

\sqrt(3) r₂ = 1.5 90°

\sqrt(3) r₂ = (3)/(2)

r₂ = \sqrt(3)2 r₂ = 60°

To achieve TIR, we must have $r₂ \geq 60°$ $r_2 \ge 60^{\circ}$.

Step 2: Constraint on Front Surface

Since $r₁ + r₂ = 75°$ $r_1 + r_2 = 75^{\circ}$, the condition $r₂ \geq 60°$ $r_2 \ge 60^{\circ}$ means:

r₁ \leq 15°

Step 3: Constraint on Incident Angle

Applying Snell's law at the first interface (air to prism):

1 \cdot i = \sqrt(3) r₁

For the limit condition $r₁ = 15°$ $r_1 = 15^{\circ}$:

i = \sqrt(3) 15° = 1.732 \times 0.25 = 0.433

We know $25° = 0.43$ $\sin 25^{\circ} = 0.43$. Thus $0.433$ $0.433$ is roughly $25°$ $\sin 25^{\circ}$. So $i \leq 25°$ $i \le 25^{\circ}$ is required. Any angle less than $25°$ $25^{\circ}$ ensures $r₁ \leq 15°$ $r_1 \le 15^{\circ}$ and thus $r₂ \geq 60°$ $r_2 \ge 60^{\circ}$.

Step 4: Analyze Options

Any angle $i < 25°$ $i < 25^{\circ}$ guarantees TIR. Option (1): between $15°$ $15^{\circ}$ and $20°$ $20^{\circ}$ (Valid) Option (2): $15°$ $15^{\circ}$ (Valid) Option (4): $< 15°$ $< 15^{\circ}$ (Valid) Since this is an MCQ frame inherently reflecting all suitable subsets, (1), (2), and (4) are all physically sufficient.

Pattern Recognition

In prism TIR problems, trace backward from the critical surface constraint. Maximum $r₂$ $r_2$ means minimum $r₁$ $r_1$, which dictates maximum $i$ $i$. If $imax = 25°$ $i_{\text{max}} = 25^{\circ}$, any choice strictly bounded below $25°$ $25^{\circ}$ is physically correct.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 2

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Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Velocity Ratio to Sine

Step 2: Triangle Mapping to Tangent

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Critical Angle for TIR

Step 2: Constraint on Front Surface

Step 3: Constraint on Incident Angle

Step 4: Analyze Options

Pattern Recognition

Chapter Mix

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