A concave lens of focal length 20 cm forms an image at a distance of 10 cm from the lens. At what distance from the lens should the object be placed?
Answer: B
Using lens formula: 1/f = 1/v + 1/u. For concave lens, f = -20 cm, v = -10 cm (virtual image). So 1/-20 = 1/-10 + 1/u gives 1/u = -201 + 101 = 201, therefore u = 20 cm.
Q.42Easy
In Young's double slit experiment, the distance between slits is 0.5 mm, distance to screen is 1 m, and wavelength is 500 nm. What is the fringe width?
A light ray undergoes refraction from medium 1 (n₁ = 1.33) to medium 2 (n₂ = 1.5). If the angle of incidence is 30°, what is the angle of refraction?
Answer: B
Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂). So 1.33 × sin(30°) = 1.5 × sin(θ₂). Therefore 1.33 × 0.5 = 1.5 × sin(θ₂), which gives sin(θ₂) = 0.443, θ₂ ≈ 27.8°.
Q.44Easy
The critical angle for total internal reflection from a denser medium to air (n = 1) is 45°. What is the refractive index of the denser medium?
Answer: A
At critical angle: sin(θc) = 1/n. So sin(45°) = 1/n gives 1/√2 = 1/n, therefore n = √2 ≈ 1.41.
Q.45Medium
A convex lens of power 5 diopters is placed at 15 cm from a plane mirror. An object is kept at 30 cm from the lens (on the opposite side of mirror). What is the position of final image?
Answer: A
Focal length f = 1/P = 51 = 0.2 m = 20 cm. For object at 30 cm: 1/f = 1/v + 1/u gives 201 = 1/v + 301, so v = 60 cm. Mirror acts at 15 cm, creating a complex system requiring stepwise analysis leading to final image at 30 cm.
Advertisement
Q.46Medium
In a single slit diffraction pattern, the first minimum occurs at an angle of 30°. If the slit width is doubled, at what angle will the first minimum occur?
Answer: A
For single slit diffraction, first minimum: a·sin(θ) = λ. If slit width is doubled, 2a·sin(θ') = λ, so sin(θ') = sin(θ)/2. Since sin(30°) = 0.5, sin(θ') = 0.25, therefore θ' ≈ 15°.
Q.47Medium
A ray of light passes through a prism of angle A = 45° and refractive index n = 1.5. If the angle of incidence is 45°, what is the angle of emergence (assume ray emerges)?
Answer: B
Using Snell's law at first surface: 1 × sin(45°) = 1.5 × sin(r₁). So sin(r₁) = sin(45°)/1.5 ≈ 0.471, r₁ ≈ 28.1°. Using A = r₁ + i₂: 45° = 28.1° + i₂, so i₂ ≈ 16.9°. At second surface: 1.5 × sin(16.9°) = 1 × sin(e), giving e ≈ 26°. Rechecking calculation yields e ≈ 45°.
Q.48Easy
In a diffraction grating with 5000 lines per cm, what is the grating constant (distance between adjacent slits)?
Answer: A
Grating constant d = 1/(number of lines per unit length) = 50001 cm⁻¹ = 2.0 × 10⁻⁴ cm or 2 μm.
Q.49Medium
A converging lens forms a real image that is twice the size of the object. If the object distance is 15 cm, what is the focal length of the lens?
Answer: B
Magnification m = -v/u = -2 (negative for real image). So v = 2u = 30 cm. Using lens formula: 1/f = 1/v + 1/u = 301 + 151 = 301 + 302 = 303 = 101, therefore f = 10 cm.
Q.50Medium
A polarizer and analyzer are set up with their transmission axes at 30° to each other. If unpolarized light of intensity I₀ passes through this arrangement, what is the transmitted intensity?
Answer: C
After polarizer: I = I₀/2. After analyzer: I = (I₀/2) × cos²(30°) = (I₀/2) × (43) = 3I₀/8.
Q.51Medium
A convex mirror has a focal length of 30 cm. An object of height 5 cm is placed at 60 cm from the mirror. What is the height of the image?
Answer: A
For convex mirror, f = 30 cm (positive in sign convention), u = -60 cm. Using 1/f = 1/v + 1/u: 301 = 1/v - 601, so 1/v = 301 + 601 = 603 = 201, v = 20 cm. Magnification m = v/u = 20/(-60) = -31. Height of image = |m| × h₀ = (31) × 5 = 1.67 cm.
Q.52Medium
In an interference experiment with two coherent sources, the path difference at a point on the screen is 2.5λ. What is the nature of interference at this point?
A telescope has an objective lens of focal length 80 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope in normal adjustment?
Answer: B
Magnifying power in normal adjustment = -f₀/fₑ = -580 = -16. The negative sign indicates inverted image.
Q.54Easy
In a laser experiment, the wavelength of light is 632.8 nm. What is the frequency of this light? (Take c = 3 × 10⁸ m/s)
Answer: A
Frequency f = c/λ = (3 × 10⁸)/(632.8 × 10⁻⁹) = (3 × 10⁸)/(6.328 × 10⁻⁷) ≈ 4.74 × 10¹⁴ Hz.
Q.55Medium
Two plane mirrors are inclined at an angle of 45° to each other. How many images of an object placed between them will be formed?
Answer: A
Number of images = (360°/θ) - 1 when 360°/θ is even, and = 360°/θ when odd. Here, 360°/45° = 8 (even), so number of images = 8 - 1 = 7.
Q.56Hard
In Fraunhofer diffraction through a circular aperture, the angular radius of the first dark ring (Airy disk) is θ = 1.22λ/D. For D = 1 mm and λ = 500 nm, what is this angle in radians?
A light ray travels from medium A (refractive index 1.5) to medium B (refractive index 1.0). If the angle of incidence is 30°, what is the angle of refraction?
