Question

In a biprism experiment, the fringes are observed in the focal plane of the eye-piece at a distance of 1.2 m from the slit. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eye-piece, 90 cm from the eye-piece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.

Hint:

In fringe experiments, the fringe width is affected by the distance between the slits, the screen, and the wavelength of the light used.

Answer 1

Step 1: Formula for fringe width.
In the biprism experiment, the fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance between the screen and the source, and \( d \) is the separation between the slits.
Step 2: Given Data.
From the given data:
- The distance between the central bright band and the 20th bright band is 0.4 cm, which corresponds to 20 fringe widths. Thus, the fringe width \( \beta \) is: \[ \beta = \frac{0.4 \, \text{cm}}{20} = 0.02 \, \text{cm} = 2 \times 10^{-4} \, \text{m} \]
- The distance between the two virtual magnified images is 0.9 cm. Using the formula for magnification: \[ M = \frac{v}{u} \] where \( v \) is the image distance and \( u \) is the object distance, the magnification \( M \) is \( \frac{0.9}{0.4} = 2.25 \).
Step 3: Calculate Wavelength.
Using the relation for fringe width and magnification, we can calculate the wavelength: \[ \lambda = \frac{\beta d}{D} \] Substituting the values: \[ \lambda = \frac{2 \times 10^{-4} \times 1.2}{0.9} = 4.5 \times 10^{-5} \, \text{m} \]