A concave mirror of short aperture and radius of curvature 10cm is cut into two equal parts, slightly separated and placed in front of a monochromatic line source S of wavelength $2000 A °$ as shown in figure. Consider only interference in reflected light after first reflection from mirrors. Assume direct light from S is blocked from reaching the screen.
Assume A is the position of first order maxima on the screen. Now the point object is moved away from mirror normal to screen with constant speed, then:

Question

A concave mirror of short aperture and radius of curvature 10cm is cut into two equal parts, slightly separated and placed in front of a monochromatic line source S of wavelength $2000 A °$ as shown in figure. Consider only interference in reflected light after first reflection from mirrors. Assume direct light from S is blocked from reaching the screen.

Assume A is the position of first order maxima on the screen. Now the point object is moved away from mirror normal to screen with constant speed, then:

Infinity Learn · Accepted Answer

As the source moves away from mirror i.e. from S to S', the value of d decreases and D increases.So, fringe width β$$=$$λDd increases.Before moving object, position of first order maxima is given by $$y=$$λDd.After movement of source, the position of first order maxima also increases, means it moves above point A. Later on first minima coincides with point A , and then as time passes central bright fringe coincides with point A.Thus , intensity at point A on screen will first decrease, then increase.

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