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

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

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

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

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

Q36 jee_main_2026_28_january_morning Lens Maker's Formula

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ____.

A. $5:2$ $5:2$
B. $5:12$ $5:12$
C. $12:5$ $12:5$
D. $2:5$ $2:5$

Solution

Related Formula

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

Core Logic

Equate the magnitudes of the power of the two lenses using the Lens Maker's formula.

Step 1: Power of Biconvex Lens

For a biconvex lens ( $μ = 1.5$ $\mu = 1.5$), let the front radius be $R₁$ $R_1$ and back radius be $R₂$ $R_2$. By sign convention, $R₁ > 0$ $R_1 > 0$ and $R₂ < 0$ $R_2 < 0$.

|PA| = (1.5 - 1)((1)/(R₁) - (1)/(-R₂)) = 0.5 ((1)/(R₁) + (1)/(R₂))

Step 2: Power of Plano-Concave Lens

For a plano-concave lens ( $μ' = 1.7$ $\mu' = 1.7$), the flat surface has radius $\infty$ $\infty$. The curved surface matches the back surface of the biconvex lens, so its radius is $R₂$ $R_2$.

|PB| = (1.7 - 1)| (1)/(-R₂) - (1)/(\infty) | = (0.7)/(R₂)

Step 3: Equating Powers

Given $|PA| = |PB|$ $|P_A| = |P_B|$:

0.5 ((1)/(R₁) + (1)/(R₂)) = (0.7)/(R₂)

(0.5)/(R₁) + (0.5)/(R₂) = (0.7)/(R₂)

(0.5)/(R₁) = (0.2)/(R₂)

Step 4: Finding the Ratio

(5)/(R₁) = (2)/(R₂)

(R₁)/(R₂) = (5)/(2)

Pattern Recognition

Matching curvatures means treating the absolute value of the radius as the same variable $R₂$ $R_2$ across the two equations. Always use magnitude equations when explicitly stated 'magnitudes of power are same'.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q49 jee_main_2026_28_january_morning Lens in Liquid

A convex lens of refractive index 1.5 and focal length

f = 18 ~cm

is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is

α \times f

. The value of

α

is ____. (refractive index of water = 4/3)

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

1fmedium = ( (μL)/(μm) - 1 ) ( (1)/(R₁) - (1)/(R₂) )

Core Logic

Use the Lens Maker's formula for the lens in air, and then for the lens in water. Divide the two equations to eliminate the radius terms and solve for the focal length in water.

Step 1: Lens in Air

1fₐᵢᵣ = ( (1.5)/(1) - 1 ) ( (1)/(R₁) - (1)/(R₂) ) = 0.5 ( (1)/(R₁) - (1)/(R₂) )

Step 2: Lens in Water

1fwater = ( (1.5)/(4/3) - 1 ) ( (1)/(R₁) - (1)/(R₂) )

= ( (4.5)/(4) - 1 ) ( (1)/(R₁) - (1)/(R₂) )

= ( (0.5)/(4) ) ( (1)/(R₁) - (1)/(R₂) )

Step 3: Ratio and Difference

fwaterfₐᵢᵣ = (0.5)/(0.5 / 4) = 4

fwater = 4fₐᵢᵣ = 4f

Difference in focal length:

fwater - fₐᵢᵣ = 4f - f = 3f

Therefore, $α = 3$ $\alpha = 3$.

Pattern Recognition

Standard result to memorize: A glass lens ( $μ = 1.5$ $\mu = 1.5$) immersed in water ( $μ = 4/3$ $\mu = 4/3$) has its focal length become exactly 4 times its focal length in air.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q28 jee_main_2026_28_january_evening Lens Maker's Formula

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure.

Lens Maker's Formula diagram for Q28 - JEE Main 2026 Evening — Two planoconvex lenses joined to form a biconvex lens with distinct refractive indices.

The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of

30 cm

from the biconvex lens, the magnification of the image will be :

A. $-2$ $-2$
B. $+2$ $+2$
C. $+2.5$ $+2.5$
D. $-2.5$ $-2.5$

Solution

Related Formula

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

1fₙₑₜ = (1)/(f₁) + (1)/(f₂)

(1)/(v) - (1)/(u) = 1fₙₑₜ

m = (v)/(u)

Core Logic

The biconvex lens acts as a combination of two thin lenses in contact. For lens 1 (left): $μ₁ = 1.5$ $\mu_1 = 1.5$, $R₁ = 15 cm$ $R_{1} = 15 \text{ cm}$, flat side radius $R₂ = \infty$ $R_2 = \infty$. For lens 2 (right): $μ₂ = 1.2$ $\mu_2 = 1.2$, flat side radius $R₁ = \infty$ $R_1 = \infty$, curved side radius $R₂ = -12 cm$ $R_2 = -12 \text{ cm}$ (applying sign convention).

Step 1: Calculate Focal Lengths

(1)/(f₁) = (1.5 - 1) ((1)/(15) - (1)/(\infty)) = 0.5 \times (1)/(15) = (1)/(30)

(1)/(f₂) = (1.2 - 1) ((1)/(\infty) - (1)/(-12)) = 0.2 \times (1)/(12) = (1)/(60)

Step 2: Calculate Equivalent Focal Length

1fₙₑₜ = (1)/(30) + (1)/(60) = (2 + 1)/(60) = (3)/(60) = (1)/(20)

So, $fₙₑₜ = +20 cm$ $f_{\text{net}} = +20 \text{ cm}$.

Step 3: Image Distance and Magnification

Given object distance $u = -30 cm$ $u = -30 \text{ cm}$.

(1)/(v) - (1)/(-30) = (1)/(20)

(1)/(v) = (1)/(20) - (1)/(30) = (3 - 2)/(60) = (1)/(60)

v = +60 cm

Magnification $m = (v)/(u) = (60)/(-30) = -2$ $m = \frac{v}{u} = \frac{60}{-30} = -2$.

Pattern Recognition

Combination of thin lenses in contact: simply add their powers $1/fₙₑₜ = 1/f₁ + 1/f₂$ $1/f_{net} = 1/f_1 + 1/f_2$. Apply lens maker's formula carefully with correct sign conventions for the planar boundary ( $R=\infty$ $R=\infty$).

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q43 jee_main_2026_28_january_evening Prism

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

A. $\sqrt(2) < n < 2\sqrt(2)$ $\sqrt{2} < n < 2\sqrt{2}$
B. $1 < n < 2$ $1 < n < 2$
C. $n \geq 2$ $n \geq 2$
D. $\sqrt(2) < n < 2$ $\sqrt{2} < n < 2$

Solution

Related Formula

μ = ( δ + A2) ((A)/(2))

δ = 2i - A

Core Logic

Given $δ = A$ $\delta_{\min} = A$.

μ = n = ( ((A + A)/(2)))/( ((A)/(2))) = ( A)/( (A/2))

n = (2 (A/2) (A/2))/( (A/2)) = 2 (A/2)

Since $A > 0$ $A > 0$, $(A/2) < 1$ $\cos(A/2) < 1$. Thus, $n < 2$ $n < 2$.

Step 1: Applying Physical Bounds

At minimum deviation, $i = δ + A2$ $i = \frac{\delta_{\min} + A}{2}$. Since $δ = A$ $\delta_{\min} = A$, we have $i = (2A)/(2) = A$ $i = \frac{2A}{2} = A$. For light to enter the prism, the angle of incidence must be less than $90^°$ $90^\circ$ (grazing incidence is the limit). Thus, $i < 90^° \Rightarrow A < 90^°$ $i < 90^\circ \Rightarrow A < 90^\circ$.

Step 2: Calculate Limits for n

Since $A < 90^°$ $A < 90^\circ$, we have $A/2 < 45^°$ $A/2 < 45^\circ$. The cosine function is strictly decreasing in the first quadrant, so: If $A/2 < 45^°$ $A/2 < 45^\circ$, then $(A/2) > (45^°) = 1\sqrt(2)$ .

Therefore:

n = 2 (A/2) > 2 \times 1\sqrt(2) = \sqrt(2)

Combining the limits:

\sqrt(2) < n < 2

Pattern Recognition

When $δmin = A$ $\delta_{min} = A$, $μ = 2 (A/2)$ $\mu = 2\cos(A/2)$. The extreme bounding conditions arise from $A > 0$ $A > 0$ and $i < 90^°$ $i < 90^\circ$. This guarantees the theoretical range $\sqrt(2) < μ < 2$ $\sqrt{2} < \mu < 2$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_02_april_evening Spherical Mirrors and Magnification

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of

12 cm

, at the same distance of

18 cm

from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is:

A. $1 / 2$ $1 / 2$
B. $2$ $2$
C. $3$ $3$
D. $1 / 3$ $1 / 3$

Solution

Related Formula

Mirror focal length:

f = (R)/(2)

Magnification (image size relative to object size):

m = (hᵢ)/(hₒ) = (f)/(f - u)

where $u$ $u$ is the object distance.

Core Logic

Given parameters:

Radius of curvature $R = 12 cm |f| = 6 cm$
Object distance $u = -18 cm$ $u = -18 \ \mathrm{cm}$

Let's calculate magnification for both mirrors:

For Convex Mirror:
Focal length $fconvex = +6 cm$ $f_{\text{convex}} = +6 \ \mathrm{cm}$ (using Cartesian sign convention)
Magnification:

m₁ = fconvexfconvex - u = (6)/(6 - (-18)) = (6)/(24) = (1)/(4)

Thus, image size is $(1)/(4) hₒ$ $\frac{1}{4} h_o$.

Step 1: Calculate magnification of concave mirror

For Concave Mirror:
Focal length $fconcave = -6 cm$ $f_{\text{concave}} = -6 \ \mathrm{cm}$
Magnification:

Convex and concave mirror ray diagram representations

m₂ = fconcavefconcave - u = (-6)/(-6 - (-18)) = (-6)/(12) = -(1)/(2)

Thus, image size is $(1)/(2) hₒ$ $\frac{1}{2} h_o$.

Step 2: Calculate the ratio of image sizes

Since the objects are identical (same height $hₒ$ $h_o$), the ratio of the sizes of the images is:

Ratio = |hᵢ₁||hᵢ₂| = (|m₁|)/(|m₂|) = (1/4)/(1/2) = (1)/(2)

Thus, the ratio of sizes is $1/2$ $1/2$.

Pattern Recognition

Sees: Parallel convex vs concave mirror magnification. Trap: Reversing the signs of focal lengths (Convex focal length is $+$ $+$, Concave is $-$ $-$ in standard coordinate systems). Shortcut: Use direct magnification equation $m = (f)/(f-u)$ $m = \frac{f}{f-u}$. For convex, $m = (6)/(24) = (1)/(4)$ $m = \frac{6}{24} = \frac{1}{4}$. For concave, $m = (-6)/(12) = -(1)/(2)$ $m = \frac{-6}{12} = -\frac{1}{2}$. Ratio of absolute values is $(1/4)/(1/2) = 1/2$ $\frac{1/4}{1/2} = 1/2$.

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 4

Solution

Related Formula

Core Logic

Step 1: Power of Biconvex Lens

Step 2: Power of Plano-Concave Lens

Step 3: Equating Powers

Step 4: Finding the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Lens in Air

Step 2: Lens in Water

Step 3: Ratio and Difference

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Lengths

Step 2: Calculate Equivalent Focal Length

Step 3: Image Distance and Magnification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Physical Bounds

Step 2: Calculate Limits for n

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate magnification of concave mirror

Step 2: Calculate the ratio of image sizes

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2026_23_january_evening

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