How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled?

(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...

(a) the fringe width will decrease EXPLANATION:    As the wavelength of light in water will decrease, hence the fringe width will decrease because the fringe width is proportional to the wavelength.

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...

Angular separation between fringes  βθ = λ/d where λ = wavelength, d = separation between coherent sources, βθ is independent of distance between the slits and screen; so angular separation (βθ) will remain unchanged.

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...