A student conducts an interference experiment using a laser source of monochroma

Question

A student conducts an interference experiment using a laser source of monochromatic light of wavelength = 635 nm. The light is incident on a double slit having a separation of d = 0. 04 mm. At what angle from the central maximum will the second-order maximum occur? minimum occur? The light is incident on a single slit of separation, a = 0.04 mm. At what angle will the second-order maximum occur? minimum occur? If the screen is 1.50 m from the double slits, what is the linear separation between the first-order maximum and the fourth-order maximum? The student places the double-slit film in water (n = 1.33) and illuminates it with monochromatic light of lambda_air = 635 nm. What is the angle from the central maximum of the second order maximum?

Explanation / Answer

a) For double slit interference we use,

dsin=m

Where a= slit seperation, m= order, =wavelength

i) For 2nd order maximum , m= 2

dsin=m

=sin^-1[m/d]=sin^-1[(2*635*10^-9)/(0.04*10^-3)] = 1.82 deg

ii) For 2nd order maximum , m= 1.5

dsin=m

= sin^-1[(1.5*635*10^-9)/(0.04*10^-3)] = 1.36 deg

b) For single slit interference we use,

asin=m

Where a= slit width, m= order, =wavelength

i) For 2nd order maximum , m= 2

asin=m

=sin^-1[m/a]=sin^-1[(2*635*10^-9)/(0.04*10^-3)] = 1.82 deg

ii) For 2nd order maximum , m= 1.5

dsin=m

= sin^-1[(1.5*635*10^-9)/(0.04*10^-3)] = 1.36 deg

c) y= mD/d

Where y= linear separation, D=screen distance

y = [(4-1)635*10^-9*1.50]/(0.04*10^-3) = 0.071m

d) water = air/n = (635*10^-9)/1.33 = 4.77*10^-7 m

For 2nd order maximum , m= 2

dsin=m

=sin^-1[m/d]=sin^-1[(2*4.77*10^-7)/(0.04*10^-3)] = 1.37 deg

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