In Young's double slit experiment, the distance between the two slits is `0.1 mm` and the wavelength of light used is `4 xx 10^(-7) m`. If the width of the fringe on the screen is 4 mm , the distance between screen and slit is

Options

A. `0.1 mm`
B. 1 cm
C. `0.1 cm`
D. 1 m

Answer

To find the distance between the screen and the slits in Young's double slit experiment, we can use the formula for fringe width (β):

\[ \beta = \frac{\lambda D}{d} \]

Where:
- \(\beta\) = fringe width
- \(\lambda\) = wavelength of light
- \(D\) = distance between the screen and the slits
- \(d\) = distance between the two slits

Given:
- \(d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m}\)
- \(\lambda = 4 \times 10^{-7} \, \text{m}\)
- \(\beta = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}\)

Now, we can rearrange the formula to solve for \(D\):

\[ D = \frac{\beta d}{\lambda} \]

Substituting the known values into the equation:

\[ D = \frac{(4 \times 10^{-3} \, \text{m})(1 \times 10^{-4} \, \text{m})}{4 \times 10^{-7} \, \text{m}} \]

Calculating the numerator:

\[ 4 \times 10^{-3} \times 1 \times 10^{-4} = 4 \times 10^{-7} \, \text{m}^2 \]

Now substituting back into the equation for \(D\):

\[ D = \frac{4 \times 10^{-7}}{4 \times 10^{-7}} = 1 \, \text{m} \]

Thus, the distance between the screen and the slits is:

\[ \boxed{1 \, \text{m}} \]