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In this lecture you will get the knowledge of the YDSE in case of oblique incidence of light.

Kartikey Pandey
IIT BHU(B.Tech)| Super 30 Physics Faculty | FOUNDER of New era online coaching. (1,76,000+ Subscribers)

Unacademy user
Sir plz recommend study material for paper 1.

  2. Screen 2 dsin 152

  3. YDSE WITH OBLIQUE INCIDENCE In YDSE, ray is incident on the slit at an inclination of e to the axis of symmetry of the experimental set-up for points above the central point on the screen, (say for P Screen d sine+dsin, (Id<< D) and for points below O on the screen, (say for P2) (if d<<D) We obtain central maxima at a point where, Ap 0. dsinGS (d sin ,-d sin%) or 0 2 0 This corresponds to the point O in the diagram Hence we have finally for path difference. d(sin 0 + sin )-for points above O d(sin 0-sin )-for points between O & O' d(sine-sin%)-for points below O'

  4. Solved Examples In YDSE with D 1m, d 1mm, light of wavelength 500 nm is incident at an angle of 0.57 w.r.t. the axis of symmetry of the experimental set up. If centre of symmetry of screen is O as shown.0 (i) find the position of central maxima () Intensity at point O in terms of intensity of central maxima lo (ii) Number of maxima lying between O and the central maxima. 0.57 ,-0.570 0.57 ys-D tan -D: _ 1 meter x 57-rad y1cm. (ii) for point 0, :0

  5. Hence, ap-d sin ,-d ,-1 mmx (10-2 rad) 10,000 nm 20 x (500 nm) Hence point O corresponds to 20th maxima intensity at O = L 19 maxima lie between central maxima and O, excluding maxima at O and central maxima.