(a) Explain why multiple intensity maxima occur on the viewing screen. (b) Figur
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Question
(a) Explain why multiple intensity maxima occur on the viewing screen.
(b) Figure shows the pattern for a double slit with a slit separation of 0.400 mm. What wavelength is used to illuminate the slit?
(c) Consider the double slit that was used to create the pattern in the previous section. Describe how the pattern would change if extra slits were added (keeping a constant slit spacing). Ensure you discuss the intensity, location and shape of the peaks.
d) Consider viewing a distant point source of monochromatic light. Ideally the image formed on the retina of the eye would be an extremely small spot. Discuss how diffraction by the aperture of the eye limits the sharpness of the spot formed on the retina. Estimate the size of this spot.
0.8 0.6 0.4 0.2 0 -5 4 3 21 2 4 Distance (mm)Explanation / Answer
The two slits between the path of the monochromatic light wave creates a phase or path difference between the light emanated for the two slits. The path difference is d sin(x).
where d is the distance between slits, x is the angle at which the light comes out of the slit. When the path difference equal to (n ) where is the wavelength and n is any integer (n=0,1,2...) , constructive interference occur. The maxima of intensity on the screen is a result of constructive interference. Since this can happen for any integer, n, there are multiple such maxima.
b) Wavelength, = zd/(n L) where z is the fringe spacing = 1.5 mm for (n=1) maxima. d = 0.44 mm, L = 1 m = 1000 mm. Therefore =0.0006 mm = 6 * 10^-4 mm = 600 nm. (nm = nanometer = 10^(-9) m)
c) The intensity vs position light curve for multiple slits is obtained by multiplying the multiple slit interference expression times the single slit diffraction expression. The grating intensity peak is proportional to the square of the number of slits illuminated.Hence, increasing the number of slits not only makes the diffraction sharper but also more intense. The positions of the maxima and minima will remain the same but some maxima peaks will be more intense while other maxima will be suppressed.
d) The light from a distant point source after entering the eyeball is diffracted by the pupil. Hence the image quality depends on the size of pupil or aperture. The angle the central maximum makes from center to edge with respect to the pupil is
x= 1.22 /d
where d is the size of the pupil. This means the human eye cannot resolve points if the angular separation between their image on the cornea smaller than x. Thus the diffraction by the pupil limits the sharpness resolution of the eye.
The size of the image spot on the cornea is inversely proportional to the pupil size.
The size of the spot is roughly L /d where L is the length between pupil and cornea = 17mm, d=diamter of the pupil = 2 mm, =0.0006 mm. Then the size of the spot is 0.0051 mm.
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