Name: Light Absorption, Reflection, and Transmission
Uploaded: 2019-07-19
Duration: 4 min 54 s
Description: When a light wave is an incident on an object, a number of things can happen. It is either absorbed, transmitted or reflected. Know more about it in this video.

Question

The distance between two consecutive bright fringes in a biprism experiment using light of wavelength

6000 Å

is

0.32 mm

by how much will the distance change if light of wavelength

4800 Å

is used?

Accepted Answer

Case $1$

Wavelength , $λ_{1} = 6000 Å = 6000 \times 10^{- 7} m$

Distance, $W_{1} = 0.32 mm = 3.2 \times 10^{- 4}$

Case $2$

Wavelength in the second case, $λ_{2} = 4800 Å = 4800 \times 10^{- 7} m$

Let the distance be $W_{2}$

Distance between consecutive bright fringes, $W = \frac{λ D}{d}$

$∴ W_{1} = \frac{6000 \times 10^{- 7} \times D}{d} - - - - - (1) ∴ W_{1} = \frac{4800 \times 10^{- 7} \times D}{d} - - - - - (2)$

$\frac{W_{1}}{W_{2}} = \frac{\frac{λ_{1} D}{d}}{\frac{λ_{2} D}{d}} = \frac{λ_{1}}{λ_{2}} \frac{3.2 \times 10^{- 4}}{W_{2}} = \frac{6000 \times 10^{- 7}}{4800 \times 10^{- 7}} W_{2} = \frac{3.2 \times 10^{- 4} \times 4800 \times 10^{- 7}}{6000 \times 10^{- 7}} W_{2} = \frac{3.2 \times 10^{- 4} \times 48}{60} = 2.56 \times \times 10^{- 4} ∆ W = W_{1} - W_{2} ∆ W = 3.2 \times 10^{- 4} - 2.56 \times 10^{- 4} ∆ W = 0.64 \times 10^{- 4} = 6.4 \times 10^{- 5} m$

The change in distance between the fringes is $0.064 mm .$ .

The distance between two consecutive bright fringes in a biprism experiment using light of wavelength $6000 Å$ is $0.32 mm$ by how much will the distance change if light of wavelength $4800 Å$ is used?

Important Questions on Wave Optics

Learn and Prepare for any exam you want !

The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000Å is 0.32 mm by how much will the distance change if light of wavelength 4800Å is used?

Important Questions on Wave Optics

Learn and Prepare for any exam you want !

The distance between two consecutive bright fringes in a biprism experiment using light of wavelength $6000 Å$ is $0.32 mm$ by how much will the distance change if light of wavelength $4800 Å$ is used?