Question

In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30°. Then θ of the first secondary maximum is:-
1. \({\tan ^{ - 1}}\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r - 1}^n \frac{2}{{r\left( {r + 1} \right)}}\)
2. 60°
3. \({\sin ^{ - 1}}\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r - 1}^n \left( {\frac{2}{r} - \frac{2}{{r\left( {r + 1} \right)}}} \right)\)
4. \({\tan ^{ - 1}}\mathop {2\lim }\limits_{n \to \infty } \left( {1 - \frac{1}{2}} \right)\)

Answer 1

Correct Answer - Option 3 : \({\sin ^{ - 1}}\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r - 1}^n \left( {\frac{2}{r} - \frac{2}{{r\left( {r + 1} \right)}}} \right)\)

n = 1, q = 30°

a sin = nl

a sin 30 = 1λ ⇒ a = 2λ.     ...(1)

For 1^st secondary maximum

a sin θ = 3 d/2

\(\left( {\frac{1}{2} - \frac{1}{3}} \right)\)      ...(2)

putting the value of a in eq (1) and eq (2)

\(\frac{{p_1^2}}{{{a^2}}} + \frac{{p_2^2}}{{{b^2}}}\)

\(\sqrt {10} \)

\(\sqrt 2 \)

Therefore, the correct answer is (C).