A single slit diffraction pattern is formed with white light. For what wavelength of light the 4 th secondary maximum in diffraction pattern coincides with the 3 rd secondary maximum in the pattern of

Question

A single slit diffraction pattern is formed with white light. For what wavelength of light the 4 th secondary maximum in diffraction pattern coincides with the 3 rd secondary maximum in the pattern of light of wavelength ' λ ' ?

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 $4^{th}$ secondary maximum of wavelength $λ^{'}$ , we substitute $n = 4$ :

$a sin θ_{4} = (2 (4) + 1) \frac{λ ^{'}}{2} = \frac{9 λ ^{'}}{2}$

For the $3^{th}$ secondary maximum of wavelength $λ$ , we substitute $n = 3$ :

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

Since the two maxima coincide, their angles of diffraction must be equal ( $θ_{4} = θ_{3}$ ). Equating the two expressions, we get:

$\frac{9 λ ^{'}}{2} = \frac{7 λ}{2}$

$9 λ^{'} = 7 λ$

$λ^{'} = \frac{7 λ}{9}$

Answer: $\frac{7 λ}{9}$

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