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4 years ago

11

A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the

axis in terms of the intensity of the central maximum.

1 answer:

MArishka [77]4 years ago

6 0

Answer:

The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is

$I_c = \frac{I}{I_o} =8.48 *10^{-8}$

Explanation:

From the question we are told that

The width of the slit is $D = 0.3 \ mm = 0.3 *10^{-3} \ m$

The wavelength is $\lambda = 254 \ nm = 254 *10^{-9} \ m$

The angle is $\theta = 11^o$

The intensity of at $11^o$ to the axis in terms of the intensity of the central maximum. is mathematically represented as

$I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2$

Where $\beta$ is mathematically represented as

$\beta = \frac{D sin (\theta ) * \pi}{\lambda }$

substituting values

$\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }$

$\beta = 708.1 \ rad$

So

$I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2$

$I_c = \frac{I}{I_o} =8.48 *10^{-8}$

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