In a single slit diffraction experiment, for wavelength ' λ ', half angular width of the principal maxima is ' θ '. Also for wavelength of light ' pλ ', the half angular width of the principal maxima is ' qθ '. The ratio of the half angular widths of the first secondary maxima in the first case to second case will be (a) p: 1 (b) 1: q (c) p: q (d) q: p

Answer: (b). Half angular width of central maximum = (λ)/(a); each secondary maximum has angular width (λ)/(a) too, so its half width is (λ)/(2a). Both widths scale the same way with λ, so secondary-maximum width ∝ principal half width. Case 1: case 2 = θ: qθ = 1: q So the answer is (b) 1: q.

In a single slit diffraction experiment, for wavelength ' λ ', half… | MHT CET 2026 (17 Apr, Shift 1)

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Diffraction at a Single Slit

In a single slit diffraction experiment, for wavelength ' $\lambda$ ', half angular width of the principal maxima is ' $\theta$ '. Also for wavelength of light ' $p\lambda$ ', the half angular width of the principal maxima is ' $q\theta$ '. The ratio of the half angular widths of the first secondary maxima in the first case to second case will be

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Answer: (b)

Half angular width of central maximum $= \dfrac{\lambda}{a}$ ; each secondary maximum has angular width $\dfrac{\lambda}{a}$ too, so its half width is $\dfrac{\lambda}{2a}$ .
Both widths scale the same way with $\lambda$ , so secondary-maximum width $\propto$ principal half width.
Case 1 : case 2 $= \theta : q\theta = 1 : q$
So the answer is (b) $1 : q$ .

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