5 In a singleslit experiment the slit width is 215 times the

5.
In a single-slit experiment, the slit width is 215 times the wavelength of the light. What is the width of the central maximum on a screen 2.2 m behind the slit?

6.

The figure below shows the light intensity on a screen 2.4 m behind an aperture. The aperture is illuminated with light of wavelength 590 nm.
(a) Is the aperture a single slit or a double slit?
single

double   

(b) If the aperture is a single slit, what is its width? If it is a double slit, what is the spacing between the slits?
7.

You need to use your cell phone, which broadcasts an 840 MHz signal, but you\'re behind two massive, radio-wave-absorbing buildings that have only a 15 m space between them. What is the angular width of the electromagnetic wave after it emerges from between the buildings?

8.

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.2 cm. You have a helium-neon laser ( = 633 nm) and a 0.12 mm diameter pinhole. How far behind the pinhole should you place the viewing screen?

Solution

slit width, a = 215*lambda

and D = 2.2 m

suppose first dark fring forms at distance y from central point.

then width of central maximum = 2y

and for dark fringe,

y = 0.5 * lambda *D / a

y = (0.5 x lambda x 2.2) / (215 * lambda )

y = 5.12 x 10^-3 m

width = 2y = 0.0102 m Or 1.02 cm OR 10.2 mm

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Figure missing

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7. suppose first dark fring forms at distance y from central point.

then width of central maximum = 2y

and for dark fringe,

sin@ = 0.5 * lambda / a

lambda = c / f = (3 x 10^8 ) / (840 x 10^6) = 0.357 m

sin@ = 0.5 x 0.357 / 15

@ = 0.682 deg

angular width = 2@ = 1.36 deg

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8. y = r = 0.5 * lambda *D / a

(1.2 x 10^-2 / 2) = (0.5 x 633 x 10^-9 x D ) / (0.12 x 10^-3)

D = 2.27 m