Question

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

Answer 1

Given :-

• The distance between two consecutive bright fringes in a baptism experiment using light of wavelength 6000 A⁰is 0.32mm

To Find :-

• how much will the distance change if light of wavelength 4800A⁰is used?

Solution :-

We know that

$\sf \beta _{i} \propto \lambda_{i}\; and \; \beta_{f}\propto \lambda_{f}$

$\sf \dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{6000 \times 10^{-10}}{4800 \times 10^{-10}}$

$\sf \dfrac{0.32\times10^{-3}}{\beta_{f}} = \dfrac{6000}{4800}$

$\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{60}{48}$

$\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}}$

$\sf 0.32\times 10^{-3} = 1.25\times\beta_{f}$

$\sf \dfrac{0.32\times 10^{-3}}{1.25}=\beta_{f}$

$\sf 0.25 \times 10^{-3}=\beta_{f}$

Now

$\sf Difference = 0.32 \times 10^{-3}-0.25\times 10^{-3}$

$\sf Difference = 0.7\times 10^{-3}$