Wave Optics Tes...

In YDSE fringe width is $$\beta$$. If distance between the slits is halved and the distance of the screen from the slit is made three times, the new fringe width will be :-

Light of wavelength 550 nm falls normally on a slit of width 22.0$$\times 10^{-5}$$ cm. The angular potion of the second minimum from the centre maximum will be ( in radians)

After reflection from a concave mirror, a plane wave front becomes 

A slit of size $$0.15cm$$ is placed at $$2.1m$$ from a screen. On illuminated it by a light of wavelength $$5 \times {10^{ - 5}}cm$$. The width of diffraction pattern will be:-

The distances of interference point on a screen from two slits are $$18 \, \mu \, m $$ and 12.3 $$\mu$$ m. If the wavelength of light used is $$6 \times 10^{-7}m $$ then the number of dark or bright fringe formed there will be - 

The displacement of the interfering sound waves are $${ y }_{ 1 }=4sincot$$ and $${ y }_{ 2 }=3sin\left( cot+\frac { \pi  }{ 2 }  \right) $$. What is the amplitude of the resultant wave

Two beams of light will not give rise to an interference pattern, if 

In an interference pattern of two waves fringe width is $$\beta$$. If the frequency of source is doubled then fringe width will become:-

In $$Y.D.S.E$$. If the width of the slits are gradually decreased, then 

Which of the following statement is/ are correct?
I. A person suffering from myopia can see distant objects clearly.
II.. The bluish color of water in deep sea is only due to reflection of sky in water.
III. In vacuum, all the colored of the light move with the same speed.
IV. At noon, the sun appears white as light is least scattered