If Young's double slit experiment is performed in water, (a) the fringe width will decrease

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half...

(a)  Straight line fringes are formed on screen.

The correct answer is (b) IO/4 EXPLANATION:  As in the previous explanation the maximum intensity  I₀ = (√I₁ + √I₂)². Here given that I₁ = I₂ = I (say).  →I₀ = 4 I  -------...

(c) remain same EXPLANATION:    The fringe width is given as w = Dλ/d where D = distance of the screen from the slits d = distance of slits.  The wavelength λ remains the same...

(b) consecutive fringes will come closer EXPLANATION:    The options (a), (c) and (d) are related to the intensity changes. Since the intensity of light incident on the slits is not changed...

The correct answer is (3) Straight line Note: If instead of Young's double slit experiment, young’s double hole experiment was given shape would have been hyperbola.

Fringe width, β = Dλ/d => β ∝  λ  for same D and d. When the whole apparatus is immersed in a  transparent liquid of refractive index n =1×3, the wavelength decreases to  ...

 (i) Angular separation βθ = β/D = λ/d It is independent of D; therefore, angular separation remains unchanged if screen is moved away from the slits. But the actual separation between fringes β = λD/d  increases, so visibility...

In single slit diffraction experiment fringe width is β = 2λD/d If d is doubled, the width of central maxima is halved. Thus size of central maxima is reduced to half. Intensity...

The fringe width is β = λD/d If D (distance between slits and screen) is doubled, then fringe width will be doubled.