First slide

Diffraction of light-Wave Optics

Question

In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength $λ$ '. The value of $λ$ ' is

Moderate

A

$4400 Å$
B

$6600 Å$
C

$2000 Å$
D

$3500 Å$

Solution

In a single slit diffraction experiment, position of minima is given by $dsinθ = nλ$
So for first minima of red $sinθ = 1 \times (\frac{λ_{R}}{d})$
and as first maxima is midway between first and second minima, for wavelength $λ'$ its position will be
$dsinθ' = \frac{λ' + 2 λ'}{2} \Rightarrow sinθ' = \frac{3 λ'}{2 d}$
According to given condition $sinθ = sinθ'$
$\Rightarrow λ' = \frac{2}{3} λ_{R} so λ' = \frac{2}{3} \times 6600 = 440 nm = 4400 \overset{o}{A}$

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