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

Q20 jee_main_2025_28_jan_morning Prism and Dispersion

A thin prism

P₁

with angle

4°

made of glass having refractive index 1.54, is combined with another thin prism

P₂

made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism

P₂

in degrees is

A. 4
B. 3
C. 16/3
D. 1.5

Solution

Related Formula

δ = (μ - 1)A

Core Logic

To achieve dispersion without deviation, the net deviation produced by the prism combination must be zero:

δₙₑₜ = 0 (μ₁ - 1)A₁ - (μ₂ - 1)A₂ = 0

Substituting the given parameters into the equation:

(1.54 - 1) \cdot 4° - (1.72 - 1)A₂ = 0

0.54 \cdot 4 = 0.72 \cdot A₂

A₂ = (2.16)/(0.72) = 3°

Step 1: Final Angle Value

The required angle for the second thin prism is $3°$ $3^{\circ}$, which matches option (2).

Pattern Recognition

For zero deviation conditions using thin components, balance the deviation equations directly: $(μ-1)A = (μ'-1)A'$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q18 jee_main_2025_03_april_morning Lens Maker's Formula

The radii of curvature for a thin convex lens are

10~cm

and

15~cm

respectively. The focal length of the lens is

12~cm

. The refractive index of the lens material is:

A. $1.2$ $1.2$
B. $1.4$ $1.4$
C. $1.5$ $1.5$
D. $1.8$ $1.8$

Solution

Related Formula

Lens Maker's Formula:

(1)/(f) = (μ - 1) ((1)/(R₁) - (1)/(R₂))

where, $f$ $f$ = focal length of the lens, $μ$ $\mu$ = refractive index of the material, $R₁, R₂$ $R_1, R_2$ = radii of curvature with standard Cartesian sign convention.

Core Logic

For a thin bi-convex lens, using standard coordinate conventions:

$R₁ = +10~cm$ $R_1 = +10\mathrm{~cm}$ (positive since first surface centers to the right of light trajectory),
$R₂ = -15~cm$ $R_2 = -15\mathrm{~cm}$ (negative since second surface centers to the left),
Focal length $f = +12~cm$ $f = +12\mathrm{~cm}$.

Step 1: Substituting in the Equation

Substitute the values into Lens Maker's formula:

(1)/(12) = (μ - 1) ((1)/(10) - (1)/(-15))

(1)/(12) = (μ - 1) ((1)/(10) + (1)/(15))

(1)/(12) = (μ - 1) ((3 + 2)/(30))

(1)/(12) = (μ - 1) ((5)/(30)) = (μ - 1) ((1)/(6))

μ - 1 = (6)/(12) = 0.5 μ = 1.5

Pattern Recognition

Convex lenses always have opposite signs for $R₁$ $R_1$ and $R₂$ $R_2$. The term $((1)/(R₁) - (1)/(R₂))$ is additive: $((1)/(|R₁|) + (1)/(|R₂|))$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q20 jee_main_2025_03_april_morning 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 $A$ $A$:

Deviation: $δ = i + e - A$ $\delta = i + e - A$
Minimum deviation $δmin$ $\delta_{\text{min}}$ occurs when:

i = e and r₁ = r₂ = (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:

μ = ( A+δmin2) ((A)/(2))

As $A$ $A$ increases, $δmin$ $\delta_{\text{min}}$ 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 $δ-i$ $\delta-i$ curve is asymmetric and parabolic-like; for any deviation $δ > δmin$ $\delta > \delta_{\text{min}}$, 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 ( $δ-i$ $\delta-i$) 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 $r₁ = r₂ = A/2$ $r_1 = r_2 = A/2$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments: Refraction through Prism

Q19 jee_main_2025_04_april_evening Spherical Mirrors

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

A. 45 cm
B. 7.5 cm
C. 22.5 cm
D. 15 cm

Solution

Related Formula

Mirror Formula:

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

Core Logic

For the convex mirror, $u = -30 cm$ $u = -30\text{ cm}$ and $f = +30 cm$ $f = +30\text{ cm}$.

(1)/(v) - (1)/(30) = (1)/(30) (1)/(v) = (2)/(30) v = +15 cm

So, the convex mirror forms a virtual image 15 cm behind its surface.

Step 1: Align Plane Mirror Image

The total distance from the object to the image location is $30 + 15 = 45 cm$ $30 + 15 = 45\text{ cm}$. For a plane mirror to create an image at this same exact location, it must be placed precisely midway between the object and the image. Distance from object to plane mirror:

d = (45)/(2) = 22.5 cm

Therefore, the clearance distance between the convex mirror and the plane mirror surface is:

Distance = 30 - 22.5 = 7.5 cm

Pattern Recognition

Coinciding images imply identical coordinate endpoints. Calculate the convex position explicitly, find the total path length from the real source object, and slice it in half for the plane mirror location.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_04_april_morning Spherical Mirrors and Magnification

When an object is placed

40~cm

away from a spherical mirror an image of magnification

(1)/(2)

is produced. To obtain an image with magnification of

(1)/(3)

, the object is to be moved:

A. 40 cm away from the mirror.
B. 80 cm away from the mirror.
C. 20 cm towards the mirror.
D. 20 cm away from the mirror.

Solution

Related Formula

Magnification formula for a spherical mirror in terms of focal length $f$ $f$ and object distance $u$ $u$:

m = (f)/(f - u)

Core Logic

Case 1: Given $u₁ = -40~cm$ $u_1 = -40\mathrm{~cm}$ and $m₁ = (1)/(2)$ $m_1 = \frac{1}{2}$:

(1)/(2) = (f)/(f - (-40)) f + 40 = 2f f = +40~cm

Since $f > 0$ $f > 0$, the optical element is a convex mirror forming a virtual, erect, and diminished image ( $0 < m < 1$ $0 < m < 1$).

Step 1: Calculate New Object Position

Case 2: To achieve a magnification of $m₂ = (1)/(3)$ $m_2 = \frac{1}{3}$:

(1)/(3) = (40)/(40 - u₂) 40 - u₂ = 120 u₂ = -80~cm

Step 2: Determine Distance Shift

Initial object position: $u₁ = -40~cm$ $u_1 = -40\mathrm{~cm}$
Final object position: $u₂ = -80~cm$ $u_2 = -80\mathrm{~cm}$

Shift = |u₂| - |u₁| = 80 - 40 = 40~cm away from the mirror

Pattern Recognition

For a convex mirror, images are always virtual, erect, and diminished. As the object is shifted farther away from the mirror ( $u \to -\infty$ $u \to -\infty$), the magnification decreases toward zero.

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 8

Solution

Related Formula

Core Logic

Step 1: Final Angle Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substituting in the Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Align Plane Mirror Image

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate New Object Position

Step 2: Determine Distance Shift

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2025_28_jan_morning

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