A plano-convex lens (n = 1.5) has radius of curvature R for the curved surface. What is its focal length?
Answer: B
Using lens maker's formula: 1/f = (n-1)[1/R₁ - 1/R₂]. For plano-convex: 1/f = (1.5-1)[1/R - 1/∞] = 0.5/R. Therefore f = 2R.
Q.502Easy
Two coherent sources of light have a phase difference of π/2 radians. What is the nature of interference at their meeting point?
Answer: C
For constructive interference, phase difference = 0, 2π, etc. For destructive interference, phase difference = π, 3π, etc. Phase difference of π/2 gives partial or incomplete interference with intermediate intensity.
Q.503Medium
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.504Medium
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.505Medium
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.
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Q.506Medium
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.507Medium
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.509Medium
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.510Medium
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.511Medium
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.513Hard
In Fraunhofer diffraction by a single slit, if the slit width is doubled, how does the angular width of the central maximum change?
Answer: B
Angular width of central maximum = 2λ/a. If slit width a doubles, angular width becomes 2λ/(2a) = λ/a, which is half the original.
Q.514Hard
Two slits of widths w₁ and w₂ produce a diffraction pattern with intensity ratio I₁:I₂ = 4:1. What is the ratio of their widths?
Answer: A
Intensity is proportional to (slit width)². If I₁:I₂ = 4:1, then w₁:w₂ = √4:√1 = 2:1.
Q.515Hard
A biconvex lens (n = 1.5) has both radii of curvature equal to 20 cm. What is its focal length?
Answer: B
Using lens maker's formula: 1/f = (n-1)[1/R₁ + 1/R₂] = (0.5)[201 + 201] = (0.5)(202) = 201. Therefore f = 20 cm.
Q.516Hard
In a double-slit experiment, if one slit is covered with a transparent film of thickness t and refractive index n, the central bright fringe shifts. What is the path difference introduced?
Answer: A
Optical path through film = nt. Geometric path = t. Extra optical path = nt - t = (n-1)t. This causes a phase shift equivalent to a path difference of (n-1)t.
Q.517Hard
A monochromatic light source of wavelength λ is incident on a diffraction grating with 500 lines/mm. The second-order maximum is at 30°. What is the wavelength?
Answer: A
Grating equation: d·sin(θ) = m·λ. Here d = 1/(500 × 10³) = 2 × 10⁻⁶ m. For m = 2: (2 × 10⁻⁶)·sin(30°) = 2·λ. (2 × 10⁻⁶)·(0.5) = 2·λ. λ = 500 nm.
Q.518Hard
A parallel beam of light undergoes diffraction through a circular aperture of diameter D. The radius of the first dark ring in Fraunhofer diffraction is proportional to:
Answer: A
For Fraunhofer diffraction by circular aperture (Airy disk), the radius of first dark ring = 1.22λf/D, which is proportional to λ/D.
Q.519Easy
A plane mirror is moved towards a stationary object at a distance of 10 cm from the mirror. What is the velocity of the image if the mirror moves with a velocity of 2 m/s towards the object?
Answer: A
When a plane mirror moves towards an object with velocity v, the image also moves towards the mirror with velocity v. The relative velocity of approach between object and image is 2v. Therefore, velocity of image = 2 × 2 = 4 m/s.
Q.520Easy
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.
