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_2026_21_jan_morning Lenses

A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.

Numerical Answer. Answer: 280 to 280

Solution

Related Formula

Magnification m = DoutDᵢₙ = (f₂)/(f₁)

Core Logic

For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.

Lenses solution diagram for Q46 - JEE Main 2026 Morning

By similar triangles at the focal point:

θ = (D₁ / 2)/(f₁) = (D₂ / 2)/(f₂)

Step 1: Solving for f2

(D₁)/(f₁) = (D₂)/(f₂)

(2)/(40) = (14)/(f₂)

f₂ = 14 \times (40)/(2) = 14 \times 20 = 280 mm

Pattern Recognition

A standard beam expander (Keplerian telescope design used in reverse) has $f₂/f₁ = D₂/D₁$ $f_2/f_1 = D_2/D_1$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q50 jee_main_2026_21_jan_morning Microscope

In a microscope the objective is having focal length

f₀ = 2 cm

and eye-piece is having focal length

fₑ = 4 cm

. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

m \approx (L)/(f₀) \times (D)/(fₑ)

Core Logic

For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation:

m l Df₀ fₑ

where $l$ $l$ is the tube length, $D = 25 cm$ $D = 25\text{ cm}$ is the least distance of distinct vision.

Step 1: Substitute Values

Given values: $l = 32 cm$ $l = 32\text{ cm}$ $f₀ = 2 cm$ $f_0 = 2\text{ cm}$ $fₑ = 4 cm$ $f_e = 4\text{ cm}$ $D = 25 cm$ $D = 25\text{ cm}$ (standard assumption when not given)

m = (32)/(2) \times (25)/(4)

m = 16 \times (25)/(4) = 4 \times 25 = 100

Microscope solution diagram for Q50 - JEE Main 2026 Morning

Pattern Recognition

Compound microscope formulas: Normal adjustment $\to$ $\to$ image at $\infty$ $\infty$, $m = (L/f₀)(D/fₑ)$ $m = (L/f_0)(D/f_e)$. Image at near point $D$ $D$, $m = (L/f₀)(1 + D/fₑ)$ $m = (L/f_0)(1 + D/f_e)$. Default to standard approximation when given tube length.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q45 jee_main_2026_21_jan_evening Prism

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

Ray diagram through prism for Q45 - JEE Main 2026 Evening — Ray of light grazing the second surface of a prism after incident parallel to the base.

If refractive index of the material of prism is

\sqrt(2)

, the angle

θ

of prism is.

A. $60°$ $60^{\circ}$
B. $75°$ $75^{\circ}$
C. $90°$ $90^{\circ}$
D. $45°$ $45^{\circ}$

Solution

Related Formula

$A = r₁ + r₂$ $A = r_1 + r_2$

μ = ( i)/( r)

Core Logic

Detailed ray angles in prism for Q45 solution - JEE Main 2026 Evening — Ray of light grazing the second surface of a prism after incident parallel to the base.

Since the incident ray is parallel to the base, looking at the geometry, the angle of incidence on the first face can be derived. The base angle is $45°$ $45^{\circ}$. Thus, the normal to the first surface makes an angle of $45°$ $45^{\circ}$ with the incident ray. Therefore, $i₁ = 45°$ $i_1 = 45^{\circ}$.

For the emergent ray to graze the second surface, the angle of emergence $e = 90°$ $e = 90^{\circ}$. Applying Snell's Law at the second surface:

μ r₂ = 1 \cdot 90°

\sqrt(2) r₂ = 1 r₂ = 1\sqrt(2)

r₂ = 45°

Step 1: First Surface Analysis

Applying Snell's Law at the incident surface:

1 \cdot (45°) = μ (r₁)

1\sqrt(2) = \sqrt(2) (r₁)

(r₁) = (1)/(2) r₁ = 30°

Step 2: Prism Geometry

The prism angle $A$ $A$ at the apex is the sum of the internal angles of refraction:

A = r₁ + r₂ = 30° + 45° = 75°

Step 3: Final Conclusion

In the triangle of the prism, the sum of angles is $180°$ $180^{\circ}$.

A + 45° + θ = 180°

75° + 45° + θ = 180°

120° + θ = 180° θ = 60°

Pattern Recognition

Grazing emergence means $r₂ = θc$ $r_2 = \theta_c$ (critical angle). Horizontal incidence with a known base angle gives $i₁$ $i_1$. Combining these through $A = r₁ + r₂$ $A = r_1 + r_2$ resolves the full geometry.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q39 jee_main_2026_22_january_morning Lens Combinations and Magnification

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification

m₁

, when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to

m₂

. The value of

| m₁m₂|

is \_\_\_\_.

A. $(5)/(9)$ $\frac{5}{9}$
B. $(5)/(27)$ $\frac{5}{27}$
C. $(3)/(2)$ $\frac{3}{2}$
D. $(25)/(27)$ $\frac{25}{27}$

Solution

Related Formula

(1)/(f) = (1)/(f₁) + (1)/(f₂) - (d)/(f₁ f₂), m = (v)/(u)

Core Logic

This question was officially dropped by the examining authority. Full marks awarded to all candidates.

Pattern Recognition

Sees: Lens combination with separation gap introduced. Shortcut: Note official exam status (Dropped question). Check: Question dropped in final key. ✓

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q40 jee_main_2026_22_january_morning Prism Refraction and Grazing Emergence

Consider an equilateral prism (refractive index

\sqrt(2)

). A ray of light is incident on its one surface at a certain angle i. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to \_\_\_\_.

A. $15°$ $15^{\circ}$
B. $20°$ $20^{\circ}$
C. $40°$ $40^{\circ}$
D. $30°$ $30^{\circ}$

Solution

Related Formula

μ r₂ = 1 e, r₁ + r₂ = A

Core Logic

Solution prism ray diagram for Q40 - JEE Main 2026 Morning

For grazing emergence ( $e = 90°$ $e = 90^{\circ}$):

μ r₂ = 1 \sqrt(2) r₂ = 1 r₂ = 45°

Since prism is equilateral ( $A = 60°$ $A = 60^{\circ}$):

r₁ = A - r₂ = 60° - 45° = 15°

Pattern Recognition

Sees: Equilateral prism with grazing emergence. Shortcut: Use critical angle condition at second surface and prism angle relation $r₁ + r₂ = A$ $r_1 + r_2 = A$. Check: Matches option (1). ✓

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

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Step 1: First Surface Analysis

Step 2: Prism Geometry

Step 3: Final Conclusion

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