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)

r1 + r2 = A Snell's Law: mu1 i = mu2 r Condition for TIR: theta > thetaC, where thetaC = murarermudenser Given prism angle A = 75. For TIR at the back surface, light travels from prism (mu = sqrt(3)) to film (mu = 1.5). Applying Snell's Law for critical angle at the back surface: To achieve TIR, we must have r2 60. Since r1 + r2 = 75, the condition r2 60 means: Applying Snell's law at the first interface (air to prism): For the limit condition r1 = 15: We know 25 = 0.43. Thus 0.433 is roughly 25. So i 25 is required. Any angle less than 25 ensures r1 15 and thus r2 60. Any angle i < 25 guarantees TIR. Option (1): between 15 and 20 (Valid) Option (2): 15 (Valid) Option (4): < 15 (Valid) Since this is an MCQ frame inherently reflecting all suitable subsets, (1), (2), and (4) are all physically sufficient. In prism TIR problems, trace backward from the critical surface constraint. Maximum r2 means minimum r1, which dictates maximum i. If imax = 25, any choice strictly bounded below 25 is physically correct. Class 12 Physics: Ray Optics and Optical Instruments

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)

r1 + r2 = A Snell's Law: mu1 i = mu2 r Condition for TIR: theta > thetaC, where thetaC = murarermudenser Given prism angle A = 75. For TIR at the back surface, light travels from prism (mu = sqrt(3)) to film (mu = 1.5). Applying Snell's Law for critical angle at the back surface: To achieve TIR, we must have r2 60. Since r1 + r2 = 75, the condition r2 60 means: Applying Snell's law at the first interface (air to prism): For the limit condition r1 = 15: We know 25 = 0.43. Thus 0.433 is roughly 25. So i 25 is required. Any angle less than 25 ensures r1 15 and thus r2 60. Any angle i < 25 guarantees TIR. Option (1): between 15 and 20 (Valid) Option (2): 15 (Valid) Option (4): < 15 (Valid) Since this is an MCQ frame inherently reflecting all suitable subsets, (1), (2), and (4) are all physically sufficient. In prism TIR problems, trace backward from the critical surface constraint. Maximum r2 means minimum r1, which dictates maximum i. If imax = 25, any choice strictly bounded below 25 is physically correct. Class 12 Physics: Ray Optics and Optical Instruments

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…

Year	2026	2025	2024	Total
Questions	19	34	10	63

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

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

Q47 jee_main_2026_23_january_evening Lenses

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ____ cm.

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

Magnification formula for thin lens in terms of $f$ $f$ and $u$ $u$:

m = (f)/(f + u)

Core Logic

Since the sizes of the images are equal at two different object positions but the focal length doesn't change, one image must be real and the other virtual. Hence, their magnifications are equal in magnitude but opposite in sign. $m₁ = -m₂$ $m_1 = -m_2$

Step 1: Set Up Magnification Equations

Object distance 1: $u₁ = -8 cm$ $u_1 = -8 \, \mathrm{cm}$ (virtual image case, closer to lens) Object distance 2: $u₂ = -24 cm$ $u_2 = -24 \, \mathrm{cm}$ (real image case)

Using proper sign convention:

(f)/(f + (-8)) = -( (f)/(f + (-24)) )

Step 2: Solve for Focal Length

(f)/(f - 8) = -(f)/(f - 24)

Assuming $f \neq 0$ $f \neq 0$:

f - 24 = -(f - 8)

$f - 24 = -f + 8$ $f - 24 = -f + 8$ $2f = 32$ $2f = 32$

f = 16 cm

Pattern Recognition

For a convex lens, if equal size images are formed at two different object distances $u₁$ $u_1$ and $u₂$ $u_2$, then $f = (u₁ + u₂)/(2)$ $f = \frac{u_1 + u_2}{2}$ if both distances are strictly positive magnitude values.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q31 jee_main_2026_24_january_morning Prism

The exit surface of a prism with refractive index n is coated with a material having refractive index

(n)/(2)

. When this prism is set for minimum angle of deviation it exactly meets the condition of critical angle. The prism angle is ____

A. $60°$ $60^{\circ}$
B. $15°$ $15^{\circ}$
C. $30°$ $30^{\circ}$
D. $45°$ $45^{\circ}$

Solution

Related Formula

Minimum Deviation: r₁ = r₂ = (A)/(2)

Critical Angle: θc = (μ₂)/(μ₁)

Core Logic

For minimum deviation, $i = e$ $i = e$ and $r = (A)/(2)$ $r = \frac{A}{2}$. At the exit surface, the light ray exactly meets the condition of the critical angle.

For Total Internal Reflection (TIR) condition matching critical angle: $r = θc$ $r = \theta_c$

r = θc

r = (n/2)/(n) = (1)/(2)

Step 1: Solve for Prism Angle

Since $r = (1)/(2)$ $\sin r = \frac{1}{2}$, we have $r = 30°$ $r = 30^{\circ}$. Since $r = (A)/(2)$ $r = \frac{A}{2}$ at minimum deviation:

(A)/(2) = 30°

$A = 60°$ $A = 60^{\circ}$

Pattern Recognition

Critical angle at an interface is always $⁻¹(μᵣₐᵣₑ/μdense)$ . If $μᵣₐᵣₑ$ $\mu_{\text{rare}}$ is exactly half of $μdense$ $\mu_{\text{dense}}$, the critical angle is exactly $30^°$ $30^\circ$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q33 jee_main_2026_24_january_morning Optical Instruments

In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is

(5)k

. The value of k is ____

A. $2$ $2$
B. $5$ $5$
C. $3.5$ $3.5$
D. $4$ $4$

Solution

Related Formula

M = ( )/(f₀) \cdot (D)/(fₑ)

Core Logic

Given parameters: Focal length of objective, $f₀ = 2 cm$ $f_0 = 2 \text{ cm}$ Focal length of eyepiece, $fₑ = 5 cm$ $f_e = 5 \text{ cm}$ Tube length, $= 10 cm$ $\ell = 10 \text{ cm}$ Least distance of distinct vision, $D = 25 cm$ $D = 25 \text{ cm}$

Step 1: Calculate Total Magnification

For normal adjustment (final image at infinity), the magnification formula is:

M = ( ( )/(f₀) ) ( (D)/(fₑ) )

M = ( (10)/(2) ) ( (25)/(5) ) = 5 \times 5 = 25

Since $M = 25 = 5²$ $M = 25 = 5^2$, and the problem states $M = (5)^k$ $M = (5)^k$, we compare to find $k=2$ $k=2$.

Pattern Recognition

For normal adjustment of a compound microscope, $M = mₒ \times mₑ \approx (L/fₒ) \times (D/fₑ)$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q28 jee_main_2026_24_january_evening Lenses and Magnification

Five persons

P₁

P₂

P₃

P₄

and

P₅

recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement

A. $Readings recorded by all persons are correct$
B. $Reading recorded by P₃ persons are incorrect$
C. $Reading recorded by P₃ and P₂ persons are incorrect$
D. $Reading recorded by P₄ and P₅ persons are incorrect$

Solution

Related Formula

P = 100f (in cm)

Core Logic

Given Power $P = +5D$ $P = +5\mathrm{D}$

(1)/(f) = 5 \Rightarrow f = 20 cm

If the object is between $f$ $f$ and $2f$ $2f$ (between $20cm$ $20\text{cm}$ and $40cm$ $40\text{cm}$), the image must be formed beyond $2f$ $2f$ (i.e., $v > 40cm$ $v > 40\text{cm}$) and it is magnified.

Step 1: Check P3 Reading

For $P₃$ $P_3$, $u = 35cm$ $u = 35\text{cm}$ (which is between $20cm$ $20\text{cm}$ and $40cm$ $40\text{cm}$). Thus, the image distance $v$ $v$ must be greater than $40cm$ $40\text{cm}$. However, the recorded reading is $v = 37cm$ $v = 37\text{cm}$, which is incorrect.

Pattern Recognition

For a convex lens, if $u$ $u$ lies between $f$ $f$ and $2f$ $2f$, $v$ $v$ must be greater than $2f$ $2f$. A simple boundary check on the $2f$ $2f$ mark ( $40cm$ $40\text{cm}$) reveals faulty readings instantly without solving the lens formula.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q45 jee_main_2026_24_january_evening Mirrors and Magnification

Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ____ cm.

A. $- 15/2$ $- 15/2$
B. $- 10$ $- 10$
C. $- 20$ $- 20$
D. $- 15$ $- 15$

Solution

Related Formula

m = -(v)/(u)

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

Core Logic

For a real image, magnification $m$ $m$ is negative: $m = -3$ $m = -3$

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

Given the distance between object and image is $40 cm$ $40\text{ cm}$: For real images formed by a concave mirror, both $u$ $u$ and $v$ $v$ are on the same side. The distance between them is $|v| - |u| = 40 cm$ $|v| - |u| = 40\text{ cm}$.

Step 1: Determine Object and Image Distances

$3|u| - |u| = 40$ $3|u| - |u| = 40$

2|u| = 40 |u| = 20 cm

|v| = 60 cm

Applying sign convention:

u = -20 cm, v = -60 cm

Step 2: Compute Focal Length

(1)/(f) = (1)/(-60) + (1)/(-20)

(1)/(f) = (-1 - 3)/(60) = -(4)/(60)

f = -15 cm

Pattern Recognition

Real magnified image means concave mirror. The separation is $(m-1)|u|$ $(m-1)|u|$. Therefore $|u| = Δ / (m-1)$ $|u| = \Delta / (m-1)$. Then strictly apply $(1)/(f) = (1)/(v) + (1)/(u)$ with signs.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Solution & Explanation

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

Reference Study Guides

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

Solution

Related Formula

Core Logic

Step 1: Set Up Magnification Equations

Step 2: Solve for Focal Length

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Prism Angle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Magnification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Check P3 Reading

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Object and Image Distances

Step 2: Compute Focal Length

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2026_23_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back