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

Q9 jee_main_2025_03_april_evening Refraction of Light and Refractive Index

A monochromatic light of frequency

5\times10¹⁴~Hz

travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

A. 300 nm
B. 600 nm
C. 400 nm
D. 500 nm

Solution

Related Formula

For light propagation, wave velocity, frequency, and wavelength are related by:

v = f λ \Rightarrow λₐᵢᵣ = (c)/(f)

When light passes into a medium of refractive index $μ$ $\mu$, the frequency remains constant, but the wavelength scales down to:

λmedium = λₐᵢᵣμ

Core Logic

Given parameters:

Frequency $f = 5 \times 10¹⁴~Hz$ $f = 5 \times 10^{14}\mathrm{~Hz}$
Speed of light in vacuum/air $c \approx 3 \times 10⁸~m/s$ $c \approx 3 \times 10^8\mathrm{~m/s}$
Refractive index of medium $μ = 2$ $\mu = 2$

Step 1: Calculate Wavelength in Air (Vacuum)

λₐᵢᵣ = 3 \times 10⁸~m/s5 \times 10¹⁴~Hz = 0.6 \times 10⁻⁶~m = 600~nm

Step 2: Calculate Refracted Wavelength in Medium

λmedium = λₐᵢᵣμ = 600~nm2 = 300~nm

Pattern Recognition

Remember: Frequency is a source characteristic and never changes during refraction. Speed and wavelength both decrease by a factor of $μ$ $\mu$ inside the medium.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_07_april_morning Refraction at Plane Surfaces

A container contains a liquid with refractive index of 1.2 up to a height of

60~cm

and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is

40~cm

. The value of H is ______

cm

. (Consider liquids are immisible)

Numerical Answer. Answer: 80 to 80

Solution

Related Formula

The apparent shift $Δ x$ $\Delta x$ in depth through multiple immiscible liquid layers viewed normally is the sum of the individual layer shifts:

Δ x = Σ dᵢ ( 1 - (1)/(μᵢ) )

Core Logic

For two liquid layers:

Layer 1: $d₁ = 60 ~cm$ $d_1 = 60 \mathrm{~cm}$, $μ₁ = 1.2$ $\mu_1 = 1.2$
Layer 2: $d₂ = H ~cm$ $d_2 = H \mathrm{~cm}$, $μ₂ = 1.6$ $\mu_2 = 1.6$

Total apparent shift is given as $Δ x = 40 ~cm$ $\Delta x = 40 \mathrm{~cm}$.

Step 1: Set Up and Solve the Equation

Substitute the parameters into the equation:

40 = 60 ( 1 - (1)/(1.2) ) + H ( 1 - (1)/(1.6) )

Calculate the fractional factors:

1 - (1)/(1.2) = 1 - (5)/(6) = (1)/(6)

1 - (1)/(1.6) = 1 - (5)/(8) = (3)/(8)

Substitute back:

40 = 60 ( (1)/(6) ) + H ( (3)/(8) )

40 = 10 + (3)/(8) H (3)/(8) H = 30

H = (30 \times 8)/(3) = 80 ~cm

Pattern Recognition Sees: Two immiscible liquid layers with normal viewing shift. Shortcut: First layer has real depth 60, index 1.2$ Pattern Recognition Sees: Two immiscible liquid layers with normal viewing shift. Shortcut: First layer has real depth 60, index 1.2 $\impliesapparent shift is$ apparent shift is $60 \times (1 - 5/6) = 10 \mathrm{~cm}. Since total shift is 40, the second layer must contribute$. Since total shift is 40, the second layer must contribute $30 \mathrm{~cm}of shift. Thus,$ of shift. Thus, $H \times (1 - 5/8) = 30 \implies H \times (3/8) = 30 \implies H = 80 \mathrm{~cm}$. Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q8 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

A lens having refractive index 1.6 has focal length of

12cm

, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

A. $355mm$ $355\mathrm{mm}$
B. $288mm$ $288\mathrm{mm}$
C. $555mm$ $555\mathrm{mm}$
D. $655mm$ $655\mathrm{mm}$

Solution

Related Formula

Lens Maker's Formula in a surrounding medium with refractive index $μm$ $\mu_m$ is:

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

Core Logic

In air ( $μm = 1$ $\mu_m = 1$):

(1)/(12) = (1.6 - 1) ( (1)/(R₁) - (1)/(R₂) )

(1)/(12) = 0.6 ( (1)/(R₁) - (1)/(R₂) ) ( (1)/(R₁) - (1)/(R₂) ) = (1)/(12 \times 0.6) = (10)/(72)

Step 1: Calculate Focal Length in Water

In water ( $μm = 1.28$ $\mu_m = 1.28$):

(1)/(fw) = ( (1.6)/(1.28) - 1 ) ( (10)/(72) )

Simplify the relative index factor:

(1.6)/(1.28) = (160)/(128) = 1.25

(1)/(fw) = (1.25 - 1) ( (10)/(72) ) = 0.25 \times (10)/(72) = (1)/(4) \times (10)/(72) = (10)/(288)

fw = 28.8 ~cm = 288 ~mm

Pattern Recognition Sees: Lens index$ Pattern Recognition Sees: Lens index $\mu_L = 1.6, focal length in air$, focal length in air $f_a, and focal length in medium$, and focal length in medium $f_m. Shortcut: Use the ratio of focal lengths directly:

(fm)/(fₐ) = (μL - 1)/((μL)/(μm) - 1) = (0.6)/((1.6)/(1.28) - 1) = (0.6)/(0.25) = 2.4

fm = 2.4 \times 12 = 28.8 ~cm = 288 ~mm

$

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q10 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

Two thin convex lenses of focal lengths

30~cm

and

10~cm

are placed coaxially,

10~cm

apart. The power of this combination is :

A. $5 D$ $5 \mathrm{D}$
B. $1 ~D$ $1 \mathrm{~D}$
C. $20D$ $20\mathrm{D}$
D. $10D$ $10\mathrm{D}$

Solution

Related Formula

The equivalent focal length $feq$ $f_{\text{eq}}$ of two thin coaxially aligned lenses separated by distance $d$ $d$ is given by:

1feq = (1)/(f₁) + (1)/(f₂) - (d)/(f₁ f₂)

The equivalent power in diopters ( $D$ $D$) when focal lengths are in meters is:

P = 1feq

Core Logic

Given parameters:

$f₁ = 30 ~cm = 0.3 ~m$
$f₂ = 10 ~cm = 0.1 ~m$
$d = 10 ~cm = 0.1 ~m$ $d = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$

Step 1: Calculate Power

Substitute parameters into the equivalent focal length equation:

1feq = (1)/(0.3) + (1)/(0.1) - (0.1)/(0.3 \times 0.1)

1feq = (1)/(0.3) + 10 - (1)/(0.3)

1feq = 10 ~m⁻¹

P = 10 ~D

Pattern Recognition

Sees: Lenses separated by distance $d$ $d$ where $d = f₂$ $d = f_2$. Shortcut: Notice that $d = f₂ = 10 ~cm$ $d = f_2 = 10 \mathrm{~cm}$. When the separation distance between two thin lenses equals the focal length of the second lens, the equivalent power simplifies directly to $P = P₂ = 1/f₂ = 10 ~D$ $P = P_2 = 1/f_2 = 10 \mathrm{~D}$ since the terms $1/f₁$ $1/f_1$ and $d/(f₁ f₂)$ $d/(f_1 f_2)$ cancel out.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_08_april_evening Refraction through Lenses

A concave-convex lens of refractive index

1.5

and the radii of curvature of its surfaces are

30~cm

and

20~cm

, respectively. The concave surface is upwards and is filled with a liquid of refractive index

1.3

. The focal length of the liquid-glass combination will be

A. $(500)/(11)~cm$ $\frac{500}{11}\mathrm{~cm}$
B. $(800)/(11)~cm$ $\frac{800}{11}\mathrm{~cm}$
C. $(700)/(11)~cm$ $\frac{700}{11}\mathrm{~cm}$
D. $(600)/(11)~cm$ $\frac{600}{11}\mathrm{~cm}$

Solution

Related Formula

1feq = 1fliquid + 1fglass

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

Core Logic

Let's find the focal length of each individual lens in the combination:

Liquid Lens:
Refractive index, $μl = 1.3$ $\mu_l = 1.3$
The upper surface is flat (exposed to air): $R₁ = \infty$ $R_1 = \infty$
The lower surface matches the upward concave surface of the glass lens: $R₂ = -30~cm$ $R_2 = -30\mathrm{~cm}$

1fliquid = (1.3 - 1) ((1)/(\infty) - (1)/(-30)) = 0.3 \times (1)/(30) = (1)/(100)~cm⁻¹

Glass Lens:
Refractive index, $μg = 1.5$ $\mu_g = 1.5$
First surface radius (concave upward), $R₁ = -30~cm$ $R_1 = -30\mathrm{~cm}$
Second surface radius (convex downward), $R₂ = -20~cm$ $R_2 = -20\mathrm{~cm}$ (following light path downward)

1fglass = (1.5 - 1) ((1)/(-30) - (1)/(-20)) = 0.5 (-(1)/(30) + (1)/(20)) = 0.5 ((1)/(60)) = (1)/(120)~cm⁻¹

Step 1: Combination Focal Length

Add the powers of both lenses:

1feq = (1)/(100) + (1)/(120) = (6 + 5)/(600) = (11)/(600)

feq = (600)/(11)~cm

Pattern Recognition

Sees: Glass lens with liquid poured on top → Think of it as a double lens system (liquid lens + glass lens). Trap: Be extremely careful with sign conventions for radii of curvature of the boundaries! Assume light travels from air through the liquid and then through the glass. This defines a consistent spatial propagation direction. ✓

Chapter Mix

Class 12 Physics: Ray Optics

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 6

Solution

Related Formula

Core Logic

Step 1: Calculate Wavelength in Air (Vacuum)

Step 2: Calculate Refracted Wavelength in Medium

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up and Solve the Equation

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Length in Water

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Power

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Combination Focal Length

Pattern Recognition

Chapter Mix

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