The figure shows a photograph of a double slit interference pattern. As explaine

Question

The figure shows a photograph of a double slit interference pattern. As explained around Figure 4 in the lab manual and in Section 35-3 of your textbook, the double-slit interference pattern has minima due to both single slit diffraction and double slit interference.

In the figure, the vertical lines indicate minima due to double silt interference. Since these minima are evenly spaced the double-slit fringes (bright spots) all have the same width, which we call yd. The two longer vertical lines indicate the width w of the central single-slit maximum. In this example, the central single-slit maximum contains 13 double slit fringes (count them).

Equation (4) in the lab manual shows the position on the screen, yd, of a double slit minimum of order n. The width yd of a fringe is equal to the difference of the positions of two consecutive minima and given by yd = L/d, where is the wavelength, L the distance to the screen, and d is the slit separation.  

In an experiment with wavelength = 633 nm and screen distance L = 1.5 m, you measured a width w = 4.9 cm for the central single-slit maximum and you counted 13 fringes.   

What is the slit separation of the double slits?

Question 4 options:

0.25 m

0.25x10-3 m

0.25x10-6 m

0.25x10-9 m

0.25 m

0.25x10-3 m

0.25x10-6 m

0.25x10-9 m

Explanation / Answer

here,

wavelength, w = 633 nm = 6.33*10^-7 m

distance to screen, D = 1.5 m

width if pattern, Y = 4.9 cm = 0.049 m

From double slit experiment we have :

slit seperation, d = (m*w*D/Y ) * no fo fringes

slit seperation, d = (6.33*10^-7)*(1.5)/0.049 * 13

slit seperation, d = 0.25 * 10^-3 m

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