Question

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 6.0 mm

Answer 1

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

  $=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

  $=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

       $=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

       $=\frac{0.31\times 6\times 10^3}{436.1}$

       $=\frac{1860}{436.1}$

       $=4.26 \ m$