Q: For a single slit, the secondary maxima occur approximately in directions given by:

a\sin\theta = (2n+1)\dfrac{\lambda}{2}. Secondary maxima are located roughly where a sinθ = (2n+1)λ/2 (n = 1, 2, …).

Question 1

Diffraction of light is the phenomenon of:

Accepted Answer

Bending of light around the edges of an obstacle or aperture. Diffraction is the bending/spreading of light as it passes the edges of an obstacle or a narrow aperture.

Question 2

Diffraction effects become significant when the size of the aperture or obstacle is:

Accepted Answer

Comparable to the wavelength of light. Appreciable diffraction occurs only when the obstacle/aperture is of the order of the wavelength of light.

Question 3

For a single slit of width a, the directions of the minima in the diffraction pattern are given by:

Accepted Answer

a\sin	heta = n\lambda (n = 1, 2, …). Single-slit minima occur where a sinθ = nλ, n = 1, 2, 3, …

Question 4

For a single slit, the secondary maxima occur approximately in directions given by:

Accepted Answer

a\sin	heta = (2n+1)\dfrac{\lambda}{2}. Secondary maxima are located roughly where a sinθ = (2n+1)λ/2 (n = 1, 2, …).

Question 5

In a single-slit diffraction pattern, the central maximum is:

Accepted Answer

The brightest and the widest. The central maximum carries most of the energy — it is the brightest and about twice as wide as the secondary maxima.

Question 6

The angular width of the central maximum in single-slit diffraction (slit width a) is:

Accepted Answer

\dfrac{2\lambda}{a}. The first minima are at sinθ = ±λ/a, so the full angular width of the central maximum is 2λ/a.

Question 7

The linear width of the central maximum on a screen at distance D from a slit of width a is:

Accepted Answer

\dfrac{2\lambda D}{a}. Linear width = 2λD/a (distance between the first minima on either side of the centre).

Question 8

In a single-slit pattern, the width of the central maximum compared with a secondary maximum is:

Accepted Answer

Twice as wide. The central maximum spans from −λ/a to +λ/a, twice the width of each secondary maximum.

Question 9

If the width a of the single slit is increased, the central maximum:

Accepted Answer

Becomes narrower. Width ∝ 1/a, so a wider slit gives a narrower (and brighter) central maximum.

Question 10

For a single slit, using light of longer wavelength makes the central maximum:

Accepted Answer

Wider. Width ∝ λ, so longer wavelength produces a wider central maximum (red spreads more than violet).

Question 11

Light of wavelength 600 nm falls on a single slit of width 0.1 mm. The linear width of the central maximum on a screen 1 m away is:

Accepted Answer

12 mm. Width = 2λD/a = (2 × 600 × 10⁻⁹ × 1)/(0.1 × 10⁻³) = 1.2 × 10⁻² m = 12 mm.

Question 12

Light of wavelength 600 nm passes through a slit of width 0.1 mm. The distance of the first minimum from the centre on a screen 1 m away is:

Accepted Answer

6 mm. First minimum: y = λD/a = (600 × 10⁻⁹ × 1)/(0.1 × 10⁻³) = 6 × 10⁻³ m = 6 mm.

Question 13

Diffraction of light provides direct evidence for the _____ nature of light.

Accepted Answer

Wave. Diffraction is a characteristic wave phenomenon, so it demonstrates the wave nature of light.

Question 14

Which is a key difference between interference and diffraction fringes?

Accepted Answer

Interference fringes are equally spaced and of equal intensity; diffraction fringes are unequally spaced with decreasing intensity. Interference gives uniform, equally spaced fringes of equal intensity, whereas diffraction gives unequally spaced fringes whose intensity falls off rapidly from the centre.

Question 15

Interference arises from the superposition of waves from _____, while diffraction arises from the superposition of secondary wavelets from _____.

Accepted Answer

two coherent sources; different parts of the same wavefront. Interference is the superposition of waves from two coherent sources; diffraction is the superposition of secondary wavelets originating from different parts of the same wavefront.

Question 16

Compared with the central maximum, the intensity of the first secondary maximum in single-slit diffraction is:

Accepted Answer

Much smaller (only a few percent). The first secondary maximum has only about 4–5% of the central maximum's intensity; the maxima fade rapidly away from the centre.

Question 17

The Fresnel distance z_F for a beam of width a and wavelength λ is:

Accepted Answer

\dfrac{a^2}{\lambda}. The Fresnel distance is z_F = a²/λ — the distance up to which a beam of width a stays approximately collimated before diffraction spreading dominates.

Question 18

The physical significance of the Fresnel distance is that it is the distance:

Accepted Answer

Up to which ray (geometric) optics is a good approximation before diffraction spreading becomes appreciable. Beyond the Fresnel distance the diffraction spread of the beam becomes comparable to its size, so ray optics is valid only up to about z_F.

Question 19

For an aperture of 3 mm and light of wavelength 500 nm, the Fresnel distance is:

Accepted Answer

18 m. z_F = a²/λ = (3 × 10⁻³)²/(500 × 10⁻⁹) = (9 × 10⁻⁶)/(5 × 10⁻⁷) = 18 m.

Question 20

If the slit width is doubled, the width of the central diffraction maximum:

Accepted Answer

Halves. Central-maximum width ∝ 1/a, so doubling a halves the width.

Diffraction — JEE Main Physics MCQs with Solutions

Questions with solutions