# A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 4.80...

## Question:

A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 4.80 m from the slits. If the bright interference fringes on the screen are separated by 1.62 cm, what is the wavelength of the laser light? nm

## Interference of light

When 2 or more waves satisfying certain conditions arrive at a point in the medium simultaneously, the resultant displacement at that point is given by the vector addition of individual displacements. When 2 or more light waves satisfying certain conditions arrive at a point in the medium simultaneously, the intensity pattern of the medium gets modified according to principle of superposition. This modification in the intensity pattern of the medium is due to the physical process called as interference of light.

Symbols Used

1) {eq}\space d. \space {/eq} is the separation between the slit

2){eq}\space D \space {/eq} is the distance between the slit and screen

3){eq}\space \lambda \space {/eq} is the wave length of light

4){eq}\space W {/eq} is the band width of interference band

• {eq}\space d =0.195 \ \rm mm \space =0.195 \times 10^{-3} \ \rm m {/eq}
• {eq}\space D = 4.8 \ \rm m {/eq}
• {eq}\space W = 1.62 \ \rm cm \space = 1.62 \times 10^{-2} \ \rm m {/eq}

Required:-

The wavelength of the laser light in nm =?

Solution:-

\begin{align} W = \frac{ \lambda \times D}{d}\\[0.3 cm] \lambda = \frac{ W \times d} {D}\\[0.3 cm] \lambda = \frac{ 1.62 \times 10^{-2} \ \rm m \times 0.195 \times 10^{-3} \ \rm m } {4.8 \ \rm m}\\[0.3 cm] \lambda = \frac{0.3159 \times 10^{-2} \ \rm m \times 10^{-3} \ \rm m} {4.8 \ \rm m }\\[0.3 cm] \lambda = \frac{0.3159 \times 10^{-5} \ \rm m^{2} } {4.8 \ \rm m }\\[0.3 cm] \lambda =0.065812 \times 10^{-5} \ \rm m \\[0.3 cm] \lambda =658.12 \times 10^{-9} \ \rm m \\[0.3 cm] \lambda =658.12 \ \rm nm \\[0.3 cm] \end{align}

The wavelength of the laser light is 658.12 nm .