Monochromatic light of wavelength 5000 A is used in Y.D.S.E., with slit-width, d 1mm, distance between screen and slits, D 1 m. If intensity at the two slits are, 1-410, 1.-10, find (i) fringe width Important (ii) distance of 5th minima from the central maxima on the screen (ii) Intensity at y mm (iv) Distance of the 1000th maxima from the central maxima on the screen. (v) Distance of the 5000th maxima from the central maxima on the screen
D 5000 x10-10x1 = 0.5 mm 1x10-3 AD 2d y = 2.25 mm y=(2n-1) d. (iii) At y = 3 mm, K< D dy 4 Now resultant intensity 1 = 11 + 12 + 2 /1112 cos = 410 + 10 + 2 V45 cos = 510 + 410 cos 3
White light is used in a YDSE with D 1 m and d = 0.9 mm. Light reaching the screen at position y 1 mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum. -9 x 104 x 1 x 10-3 m - 900 nm for minima p-(2n-1 )a/2 2AP 1800 = (2n-1) = (2-1) 1800 1800 1800 1800 3 5 7 of these 600 nm and 360 nm lie in the visible range. Hence these will be missing lines in the visible spectrum
A beam of light consisting of wavelengths 6000A and 4500A is used in a YDSE with D 1m and d 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide D 6000 x10-10 x1 d= 10-3 2D 0.45 mm Let nith maxima of 1 and nth maxima of 2 coincide at a position y. then, y n, 1-n, ,-LCM of , and 2 y = LCM of 0.6 mm and 0.45 mm y 1.8 mm Ans. At this point 3rd maxima for 6000 A & 4th maxima for 4500 A coincide
d 10-3 -0.5x10-6 n 1000 is not << 2000 Hence now p-d sin must be used Hence, d sin - --1 000 ysDtan 8275 meter (v) Highest order maxima nmax2000 Hence, n 5000 is not possible.
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