Question

A diffraction pattern is formed on a screen 130 cm away from a 0.420-mm-wide slit. Monochromatic 546.1-nm light is used. Calculate the fractional intensity I/Imax at a point on the screen 4.10 mm from the center of the principal maximum.

Answer 1

Answer:

The fractional Intensity $\frac{I}{I_{max}}$ = 0.0146

Given:

wavelength of the light, $\lambda = 546.1 nm = 546.1\times 10^{-9} m$

slit and screen separation difference, D = 130 cm = 1.3 m

distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m

slit width, d = 0.420 mm = $0.420\times 10^{-3}$

Solution:

To calculate the fractional intensity, we use the given formula:

$\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}$             (1)

$\delta = \frac{\pi }{\lambda}dsin\theta$    

For very small angle:                                        

$\delta = \frac{\pi dy}{\lambda D}$                                  (2)

where

$\delta = total phase angle$

$\theta = angle of deviation$

Using eqn (2):

$\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians$

Now, using eqn (1):

$\frac{I}{I_{max}} = \frac{sin^{2}(7.6202) }{(\frac{7.6202}{2})^{2}} = 0.0146$