Question

the distance between two consecutive bright fringes in a baptism experiment using light of wavelength 6000 A⁰is 0.32mm by how much will the distance change if light of wavelength 4800A⁰is used?​

Answer 1

Answer:

Change in distance = 0.064 mm

Explanation:

Given:

• Distance between two consecutive bright fringes = 0.32 mm = 0.32 × 10⁻³ m
• Wavelength of light used = 6000 A⁰ = 6000 × 10⁻¹⁰ m

To Find:

• Change in distance if light of wavelength 4800 A⁰ or 4800 × 10⁻¹⁰ m is used

Solution:

We know the equation,

$\sf \beta=\dfrac{D\lambda}{d}$

where β is the fringe width, D is the distance of the screen from the slit , λ is the wavelength of light used and d is the separation between the slits.

Since D and d are constants here,

β ∝ λ

Therefore,

$\sf \dfrac{\beta_i}{\beta_f} =\dfrac{\lambda_i}{\lambda_f}$

Substitute the given data,

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

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

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

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

$\sf \beta_f=0.256\times 10^{-3}$

Therefore change in distance is given by,

$\sf \beta_f-\beta_i=0.32\times 10^{-3}-0.256\times 10^{-3}$

$\sf \implies 0.064\times 10^{-3}\: m\:or\:0.064\:mm$

Hence the change in distance is 0.064 mm.

Answer 2

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

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

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

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

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

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

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

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

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

Now

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

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