Wave Optics Tes...

The YDSE apparatus is as shown in the figure. The condition for point $$P$$ to be a dark fringe is ($$l=$$wavelength of light waves)

A long horizontal slit is place $$1\ mm$$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen $$1\ m$$ away from the slit. If the mirror reflects only $$64\%$$ of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is :

If a maxima is formed at a detector, then the magnitude of wavelength $$\lambda$$ of the wave produced is given by

At $$t=0$$, fringe width is $${\beta}_{1}$$, and at $$t=2s$$, fringe width of figure is $${\beta}_{2}$$. Then

If a light of wavelength $$\lambda$$ is incident normally and the thickness of film is $$t$$, then optical path difference between waves reflected from upper and lower surface of the film is

In a Young's double slit experiment set up, source $$S$$ of wavelength $$500 nm$$ illuminates two slits $$S_{1}$$ and $$S_{2}$$ which act as two coherent sources. The source $$S$$ oscillates about its own position according to the equation $$y = 0.5 \sin \pi t$$ where $$y$$ is in mm and $$t$$ in seconds. The minimum value of time $$t$$ for which the intensity at point $$P$$ on the screen exactly infront of the upper slit becomes minimum is :

If a minima is formed at the detector, then the magnitude of wavelength $$\lambda$$ of the wave produced is given by

If $$z=\cfrac{\lambda D}{d}$$

At $$t=2s$$, the position of central maxima is

Determine the width of the region where the fringes will be visible