Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))
Q.5Medium
When white light passes through a prism, violet light deviates more than red light. This is because:
Answer: D
Higher frequency → higher refractive index → slower speed in medium → greater deviation. All statements are correct.
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Q.6Medium
An object is placed 10 cm from a concave lens of focal length 20 cm. Find the magnification.
Answer: C
Using lens formula: 1/(-20) = 101 + 1/v. v = -6.67 cm. Magnification m = -v/u = 6.1067 = 0.67
Q.7Medium
Polarization of light proves that light is:
Answer: B
Only transverse waves can be polarized. Polarization demonstrates the transverse nature of electromagnetic waves.
Q.8Medium
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:
Answer: B
Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²
Q.9Medium
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?
Answer: A
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.
Q.10Medium
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?
Answer: A
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.
Q.11Medium
A ray undergoes total internal reflection at a critical angle θc. If the refractive index of the denser medium is √2, what is θc?
Answer: B
At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.
Q.12Medium
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?
Answer: B
For concave lens, images are always virtual, erect, and diminished regardless of object position. Using 1/v = 1/f - 1/u = -201 - 101 = -203, v = -320 ≈ -6.67 cm (virtual).
Q.13Medium
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 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.15Medium
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.16Medium
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.17Medium
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.19Medium
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.20Medium
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°.
