In an interference pattern $16^{t h}$ order maximum corresponds to 600nm . What order will be visible if 480nm is used in the same position:

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Solution

The correct option is D 20
$n = 16, y = \frac{n λ D}{d} = \frac{n^{'} λ^{'} D}{d}$
$n λ = n^{'} λ^{'}$
$\Rightarrow 16 \times 600 = n^{'} \times 480$
$\Rightarrow n^{'} = 20$

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In an interference pattern 16th order maximum corresponds to 600nm . What order will be visible if 480nm is used in the same position:

The correct option is D 20n=16,y=nλDd=n′λ′Dd nλ=n′λ′ ⇒ 16×600=n′×480 ⇒ n′=20

In an interference pattern $16^{t h}$ order maximum corresponds to 600nm . What order will be visible if 480nm is used in the same position:

The correct option is D 20
$n = 16, y = \frac{n λ D}{d} = \frac{n^{'} λ^{'} D}{d}$
$n λ = n^{'} λ^{'}$
$\Rightarrow 16 \times 600 = n^{'} \times 480$
$\Rightarrow n^{'} = 20$