A single slit diffraction pattern is formed with light of wavelength 6384 A ˚ . The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of

Question

A single slit diffraction pattern is formed with light of wavelength 6384 A ˚ . The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' λ 0 ′ ​ . The value of ' λ 0 ′ ​ is

TestHub · Accepted Answer

For a single slit diffraction pattern, the condition for the $n$ -th secondary maximum is given by:

$a sin θ = (2 n + 1) \frac{λ}{2}$

For the second secondary maximum ( $n = 2$ ) of wavelength $λ$ :

$a sin θ = (2 (2) + 1) \frac{λ}{2} = \frac{5 λ}{2}$

For the third secondary maximum ( $n = 3$ ) of wavelength $λ_{0}$ :

$a sin θ = (2 (3) + 1) \frac{λ _{0}}{2} = \frac{7 λ _{0}}{2}$

Since the two maxima coincide, their angular positions are equal:

$\frac{5 λ}{2} = \frac{7 λ _{0}}{2}$

$5 λ = 7 λ_{0}$

$λ_{0} = \frac{5 λ}{7}$

Substituting $λ = 6384 \overset{˚}{A}$ :

$λ_{0} = \frac{5 \times 6384}{7} = 5 \times 912 = 4560 \overset{˚}{A}$

Answer: $4560 \overset{˚}{A}$

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