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

Q jee_main_2025_24_jan_morning Silvering 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 = 2PL + PM

(1)/(f) = 2fL + 1fM

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 (

R₂ = \infty

), meaning its mirror focal component is

fM = \infty PM = 0

. The power depends entirely on the refraction step:

(1)/(f) = 2fL

Step 1: Lens Maker Formulation

Find the focal expression of the immersed lens element [cite: 91, 679]:

1fL = ( μglassμliquid - 1)((1)/(R)) = ((1.5)/(1.2) - 1)(1)/(R) = (0.3)/(1.2)(1)/(R) = (1)/(4R)

Now insert this into the total system tracking balance relation :

(1)/(f) = 2 ((1)/(4R)) = (1)/(2R)

Given the final effective concave configuration matches $f = 0.2 m$ $f = 0.2\text{ m}$ :

(1)/(0.2) = (1)/(2R) 2R = 0.2 R = 0.10 m

Pattern Recognition

Silvering a plano-flat back boundary means light traverses the initial curved face interface exactly twice, mapping to $R = 2 \cdot f \cdot (μrel - 1)$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_24_jan_morning Lens Maker's Formula

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of

f₁

in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of

f₂

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

f₁

and

f₂

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 $μm$ $\mu_{m}$ is:

(1)/(f) = ( μlensμm - 1)( 1R₁ - 1R₂)

Core Logic

For a plano-convex configuration, the flat side has an infinite radius of curvature ( $R₂ = \infty 1R₂ = 0$ ).

Step 1: Evaluating Respective Focal Scales

For the first lens setup in air ( $μm = 1$ $\mu_{m} = 1$) :

1f₁ = (1.5 - 1)((1)/(2) - 0) = 0.5 \times (1)/(2) = (1)/(4) f₁ = 4 cm

For the second lens setup immersed inside fluid ( $μm = 1.2$ $\mu_{m} = 1.2$) :

1f₂ = ((1.5)/(1.2) - 1)((1)/(3) - 0) = (1.25 - 1)(1)/(3) = (0.25)/(3) = (1)/(12) f₂ = 12 cm

Taking their direct ratio :

f₁ : f₂ = 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.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q2 jee_main_2025_24_jan_morning Power 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 = (1)/(P)

Relative decrease in focal length is defined as:

(Δ F)/(F) = (F - F')/(F)

Core Logic

Given initial power $P = 2.5 D$ $P = 2.5\text{ D}$[cite: 19, 600]. After an increase of $0.1 D$ $0.1\text{ D}$, the new power is[cite: 19, 603]:

P' = 2.5 + 0.1 = 2.6 D

Step 1: Calculate Focal Length Change

Find the initial and final focal lengths [cite: 602, 604]:

F = (1)/(2.5) = (2)/(5)

F' = (1)/(2.6) = (5)/(13)

Now, calculate the relative decrease:

(F - F')/(F) = 1 - (F')/(F) = 1 - (P)/(P') = 1 - (2.5)/(2.6) = (0.1)/(2.6) = (1)/(26) \approx 0.04

Pattern Recognition

For a small change, we can approximate using differentiation: $P = (1)/(F) dP = -(dF)/(F²) (dF)/(F) = -(dP)/(P)$ . Thus, the relative change magnitude is $(0.1)/(2.5) = (1)/(25) = 0.04$ $\frac{0.1}{2.5} = \frac{1}{25} = 0.04$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q6 jee_main_2025_28_jan_evening Refraction at Spherical Surfaces

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at

13cm

from the vertex of the meniscus in A forms an image with a magnification of

-2

then the radius of curvature of meniscus is :

A. $1 cm$ $1 \, \text{cm}$
B. $(1)/(3)cm$ $\frac{1}{3}\mathrm{cm}$
C. $(2)/(3) ~cm$ $\frac{2}{3} \mathrm{~cm}$
D. $(4)/(3) ~cm$ $\frac{4}{3} \mathrm{~cm}$

Solution

Related Formula

For refraction at a single spherical surface separating two mediums :

(n₂)/(v) - (n₁)/(u) = (n₂ - n₁)/(R)

Linear magnification for a spherical refracting boundary is given by:

$m = (v / n₂)/(u / n₁) = (v \cdot n₁)/(u \cdot n₂)$$

Core Logic

Given parameters from the [cite: 17, 651, 654]:

Refractive index of Medium A, $n₁ = 1.3$
Refractive index of Medium B, $n₂ = 1.4$
Object distance, $u = -13 cm$
Magnification, $m = -2$

Using the magnification formula to locate image position $v$ :

-2 = \frac{v \cdot 1.3}{(-13) \cdot 1.4}

-2 = \frac{1.3 \cdot v}{-18.2} \implies 1.3 v = 36.4 \implies v = 28 \text{ cm}

Now substitute $u = -13$, $v = 28$, $n₁ = 1.3$, $n₂ = 1.4$ into the boundary equation:

\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}

\frac{1}{20} + \frac{1}{10} = \frac{0.1}{R}

\frac{1 + 2}{20} = \frac{0.1}{R} \implies \frac{3}{20} = \frac{1}{10R}

30 R = 20 \implies R = \frac{2}{3} \text{ cm}$

Refraction at Spherical Surfaces diagram for Q6 - JEE Main 2025 Evening

Q13 jee_main_2025_28_jan_evening Reflection by Spherical Mirrors

A concave mirror produces an image of an object such that the distance between the object and image is

20 cm

. If the magnification of the image is

-3'

, then the magnitude of the radius of curvature of the mirror is:

A. $3.75cm$ $3.75\mathrm{cm}$
B. $30cm$ $30\mathrm{cm}$
C. $7.5cm$ $7.5\mathrm{cm}$
D. $15cm$ $15\mathrm{cm}$

Solution

Related Formula

Magnification $m$ $m$ of a mirror is given by:

m = -(v)/(u)

The mirror equation relates focal length to distance positions:

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

Radius of curvature $R = 2f$ $R = 2f$.

Core Logic

Given, magnification $m = -3$ $m = -3$. This tells us the image is real and inverted[cite: 131, 757]:

-3 = -(v)/(u) v = 3u

Since both real objects and real images lie on the same side in front of a concave mirror, $u$ $u$ and $v$ $v$ are both negative fields. The physical distance separation between them is:

|v| - |u| = 20 3|u| - |u| = 20 2|u| = 20 |u| = 10 cm

Therefore, object distance $u = -10 cm$ $u = -10 \text{ cm}$ and image distance $v = -30 cm$ $v = -30 \text{ cm}$ .

Substitute into the focal equation formula:

f = ((-10)(-30))/(-10 - 30) = (300)/(-40) = -7.5 cm

R = 2 \times |f| = 2 \times 7.5 = 15 cm

Step 1: Visual Diagram

The visual positioning profile tracking focal path boundaries is given below:

Reflection by Spherical Mirrors diagram for Q13 - JEE Main 2025 Evening

Pattern Recognition

A magnification of $-3$ $-3$ tells you immediately that the object lies between the Focus ( $F$ $F$) and Center of Curvature ( $C$ $C$), while the image forms beyond $C$ $C$. This geometric layout instantly verifies that the radius value must exceed $10 cm$ $10\text{ cm}$.

Chapter Mix

Class 12 Physics: Ray Optics

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 10

Solution

Related Formula

Core Logic

Step 1: Lens Maker Formulation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating Respective Focal Scales

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Length Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Visual Context

Pattern Recognition

Step 1: Visual Context

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Visual Diagram

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2025_28_jan_morning

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