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?
Answer: A
Using Snell's law: sin(60°) = √3 × sin(r). √23 = √3 × sin(r). sin(r) = 21, therefore r = 30°.
Q.243Medium
A prism has apex angle A = 60° and refractive index n = √3. What is the minimum angle of deviation?
Answer: A
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°.
Q.244Medium
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?
Answer: A
At u = 30 cm: m = -f/(u-f) = -1515 = -1. At u = 20 cm: m = -515 = -3. Magnification increases in magnitude from 1 to 3.
Q.245Medium
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 lens combination has two lenses with powers P₁ = +10 D and P₂ = +5 D placed in contact. What is the focal length of the combination?
Answer: A
For lenses in contact: P_total = P₁ + P₂ = 10 + 5 = 15 D. Therefore f = 1/P = 151 ≈ 0.067 m = 6.7 cm.
Q.247Medium
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.
Q.248Medium
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.249Medium
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.250Medium
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.251Medium
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.252Medium
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.253Medium
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.255Medium
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.256Medium
In a single slit diffraction pattern, the first minima on either side of the central maximum occur at angular positions where the path difference is:
Answer: A
For single slit diffraction, minima occur when path difference = (2n+1)λ/2. For first minima (n=0), path difference = λ/2, but considering from edges, effective path difference is λ
Q.257Medium
A monochromatic light source produces a diffraction pattern using a diffraction grating. If the wavelength is decreased, the diffraction angle for the first order maximum will:
Answer: A
Using grating equation: d×sin(θ) = mλ. For fixed m and d, if λ decreases, sin(θ) decreases, hence θ decreases
Q.258Medium
An air bubble of radius 2 mm is formed in water (refractive index 1.33). The critical angle for light traveling from water to air is approximately:
