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(a) Assume that the light of wavelength 6000 Å is coming from a star. Find the limit of resolution of a telescope whose objective has a diameter of 250 cm. (b) Two slits are made 1 mm apart and the screen is placed 1m away. What should be the width of each slit to obtain 10 maxima of the double-slit patterns within the central maximum of the single-slit pattern? |
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Answer» Solution :(a) ` lambda = 6000 Å` `D = 250 cm = 250 xx 10^(-2) m` Limit of resolution = `(1.22 lambda)/D = (1.22 xx 6000 xx 10^(-10))/(250 xx 10^(-2))` `= 29.28 xx 10^(-8) m`. (B) `d = 1 MM = 1 xx 10^(-8) m` `D = 1m` Width of 10 MAXIMA of double -slit pattern `= (10 lambda D)/(d)` The width of central maxima of single-slit pattern `=(2 lambda D)/a` a is aperture of slit `:. (10 lambda D)/d = (2 lambda D)/(a) a = (2 lambda d)/(10 lambda D)` `a = (2D)/(10 D)` `= (2 xx 10^(-3))/(10)` `= 2 xx 10^(-4) m`. |
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