2001 AIPMT Wave OpticsBragg's Law of Diffraction Easy Question 40 of 40 in Wave Optics

The interplanar distance in a crystal is $2.8 \times 10^{-8}$ m. The value of maximum wavelength which can be diffracted : -

A $2.8 \times 10^{-8}$ m
B $5.6 \times 10^{-8}$ m
C $1.4 \times 10^{-8}$ m
D $7.6 \times 10^{-8}$ m

The answer is B.

Solution

Bragg's law: $2d\sin\theta = n\lambda$
Since $\sin\theta \leq 1$, for the first order (n = 1): $\lambda \leq 2d$
Therefore $\lambda_{max} = 2d = 2 \times 2.8 \times 10^{-8}$ m
$\lambda_{max} = 5.6 \times 10^{-8}$ m

Concept Maximum wavelength from Bragg's law

Wave Optics in past papers

40 questions from this chapter have appeared across 18 exam years.