In Young’s double slit experiment the distance between the slits and the screen

In Young’s double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduces to half. As a result the fringe width

Options

(a) is doubled
(b) is halved
(c) becomes four times
(d) remains unchanged

Correct Answer:

becomes four times

Explanation:

Fringe width β = Dλ/d;
From qn D’ = 2D and d’ = d/2
β’ = λD¹ / d¹ = 4β

admin:

Related Questions

  1. In R-L-C series circuit, the potential differences across each element is 20 V.
  2. A 1 kg particle strikes a wall with velocity 1 m/s at an angle 30° and reflects
  3. An electron in potentiometer experiences a force 2.4×10⁻¹⁹N. The length of potentiometer
  4. The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
  5. The ratio of the nuclear radii of elements with mass numbers 216 and 125 is