The forbidden energy gap (E_g) of germanium at 300K is approximately:

The correct answer is B: 0.66 eV. Germanium has a bandgap of approximately 0.66 eV at room temperature (300K), making it a narrow bandgap semiconductor compared to silicon (1.1 eV).

Five moles of an ideal monatomic gas are heated at constant volume from 300 K to 600 K. The change in internal energy is:

The correct answer is B: 37,440 J. ΔU = nCᵥΔT = 5 × (3/2)R × 300 = 5 × 12.5 × 8.314 × 300 = 37,440 J

Two blocks A (mass 3 kg) and B (mass 2 kg) are connected by a light string over a frictionless pulley. Block A is on a smooth horizontal surface and B hangs vertically. What is the acceleration of the system? (g = 10 m/s²)

The correct answer is A: 4 m/s². For Atwood-like system with one block horizontal: a = (m_B × g)/(m_A + m_B) = (2 × 10)/(3 + 2) = 20/5 = 4 m/s²

Two identical metal spheres with charges +Q and -Q are brought in contact and then separated. What is the final charge on each sphere?

The correct answer is B: 0 and 0. When identical conducting spheres touch, charge distributes equally. Total charge = Q - Q = 0. Each sphere gets 0 charge.

The focal length of a convex lens is 15 cm. An object is placed at 30 cm from the lens. Find the position of the image.

The correct answer is A: 30 cm (real). Using lens formula: 1/f = 1/u + 1/v. 1/15 = 1/30 + 1/v. 1/v = 1/15 - 1/30 = 1/30. v = 30 cm (real image)

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Optics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

47 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 31–40 of 47

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.31 Medium Optics

In an optical fiber, light undergoes total internal reflection. If the core has n = 1.5 and cladding has n = 1.48, what is the critical angle inside the core?

A 78.6°

B 83.4°

C 81.2°

D 85.5°

Correct Answer: B. 83.4°

EXPLANATION

sin(θc) = n_cladding/n_core = 1.48/1.5 ≈ 0.9867. Therefore θc = arcsin(0.9867) ≈ 83.4°.

Test

Q.32 Medium Optics

An object moves towards a concave mirror of focal length 15 cm. Initially at 30 cm, it moves to 20 cm. How does the magnification change?

A Increases from 1 to 3

B Decreases from 3 to 1

C Increases from 1 to 2

D Decreases from 2 to 1

Correct Answer: A. Increases from 1 to 3

EXPLANATION

At u = 30 cm: m = -f/(u-f) = -15/15 = -1. At u = 20 cm: m = -15/5 = -3. Magnification increases in magnitude from 1 to 3.

Test

Q.33 Medium Optics

A prism has apex angle A = 60° and refractive index n = √3. What is the minimum angle of deviation?

A 30°

B 45°

C 60°

D 90°

Correct Answer: A. 30°

EXPLANATION

At minimum deviation: A = r₁ + r₂ = 2r (by symmetry). Also, sin(A/2) = n·sin(r/2). sin(30°) = √3·sin(30°), which checks out. δ_m = 2i - A where i = A/2 + δ_m/2. Solving: δ_m = 30°.

Test

Q.34 Medium Optics

A ray of light is incident on a glass slab at 60°. If the refractive index of glass is √3, what is the angle of refraction?

A 30°

B 45°

C 60°

D 75°

Correct Answer: A. 30°

EXPLANATION

Using Snell's law: sin(60°) = √3 × sin(r). √3/2 = √3 × sin(r). sin(r) = 1/2, therefore r = 30°.

Test

Q.35 Medium Optics

In a Newton's rings experiment, the diameter of the 10th dark ring is 0.5 cm. What is the diameter of the 5th dark ring?

A 0.25 cm

B 0.35 cm

C 0.354 cm

D 0.4 cm

Correct Answer: C. 0.354 cm

EXPLANATION

For Newton's rings: D_m ∝ √m. Therefore D₅/D₁₀ = √5/√10 = √(1/2). D₅ = 0.5 × √(5/10) = 0.5 × √0.5 ≈ 0.354 cm.

Test

Q.36 Medium Optics

A concave lens of focal length -20 cm is used to form an image of an object placed 10 cm from it. What is the nature of the image?

A Real, inverted, diminished

B Virtual, erect, diminished

C Virtual, inverted, magnified

D Real, erect, magnified

Correct Answer: B. Virtual, erect, diminished

EXPLANATION

For concave lens, images are always virtual, erect, and diminished regardless of object position. Using 1/v = 1/f - 1/u = -1/20 - 1/10 = -3/20, v = -20/3 ≈ -6.67 cm (virtual).

Test

Q.37 Medium Optics

A ray undergoes total internal reflection at a critical angle θc. If the refractive index of the denser medium is √2, what is θc?

A 30°

B 45°

C 60°

D 75°

Correct Answer: B. 45°

EXPLANATION

At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.

Test

Q.38 Medium Optics

In Young's double-slit experiment with slit separation d = 1 mm and distance to screen D = 1 m, if the 5th bright fringe is at 2.5 mm from the center, what is the wavelength of light?

A 500 nm

B 600 nm

C 400 nm

D 550 nm

Correct Answer: A. 500 nm

EXPLANATION

For bright fringes: y = (m·λ·D)/d. For 5th bright fringe: 2.5 × 10⁻³ = (5 × λ × 1)/(1 × 10⁻³). Therefore λ = 500 nm.

Test

Q.39 Medium Optics

An object is placed at distance u from a convex lens of focal length f. If the magnification is -2, what is the relationship between u and f?

A u = 3f/2

B u = f/2

C u = 2f

D u = f/3

Correct Answer: A. u = 3f/2

EXPLANATION

Magnification m = -v/u = -2, so v = 2u. Using lens equation: 1/f = 1/u + 1/v = 1/u + 1/(2u) = 3/(2u). Therefore u = 3f/2.

Test

Q.40 Medium Optics

The intensity at a point in the interference pattern of two coherent sources is I₁ and I₂. The resultant intensity is maximum when the phase difference is:

A π

B 0 or 2π

C π/2

D 3π/2

Correct Answer: B. 0 or 2π

EXPLANATION

Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²

Test

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