Question

In Young's double slit experiment the light of 500nm is incident to obtain interference fringes of width 6nm on a screen 1 m away. 1) Calculate the diatance between two slits. 2) Find the distance of 5th bright fringe from the central bright fringe. ​

Answer 1

Given:

In Young's double slit experiment, light of 500nm is incident to obtain interference fringes of width 6nm on a screen 1 m away.

To find:

• Distance between two slits

• Distance between 5th Bright fringe and central fringe

Calculation:

In Young's Double Slit Experiment , it is assumed that the distance between two slits is much lesser as compared to the distance between slit and the screen.

Let , distance between slits be d , distance between slit and screen be D , wavelength be denoted by $\lambda$ and fringe width be denoted by $\beta$.

$\therefore \: \beta = \dfrac{ \lambda D}{d}$

$= > \: 6 \times {10}^{ - 9} = \dfrac{ 500 \times {10}^{ - 9} \times 1}{d}$

$= > \: 6 \times \cancel{ {10}^{ - 9}} = \dfrac{ 500 \times \cancel{ {10}^{ - 9}} \times 1}{d}$

$= > d = \dfrac{500}{6}$

$= > d = 83.33 \: m$

Distance of 5th bright fringe from central fringe will be:

$\therefore \: D_{n} = \dfrac{n \lambda D}{d}$

$= > \: D_{5} = \dfrac{5 \times 500 \times {10}^{ - 9} \times 1}{83.33}$

$= > D_{5} = 30 \times {10}^{ - 9} m$

$= > D_{5} = 30 \: nm$

So final answer is :

• d = 83.33 m
• $D_{5}$ = 30 nm

Answer 2

Step-by-step explanation:

Given:

In Young's double slit experiment, light of 500nm is incident to obtain interference fringes of width 6nm on a screen 1 m away.

To find:

Distance between two slits

Distance between 5th Bright fringe and central fringe

Calculation:

In Young's Double Slit Experiment , it is assumed that the distance between two slits is much lesser as compared to the distance between slit and the screen.

Let , distance between slits be d , distance between slit and screen be D , wavelength be denoted by \lambdaλ and fringe width be denoted by \betaβ .

\therefore \: \beta = \dfrac{ \lambda D}{d}∴β=

d

λD

= > \: 6 \times {10}^{ - 9} = \dfrac{ 500 \times {10}^{ - 9} \times 1}{d}=>6×10

−9

=

d

500×10

−9

×1

= > \: 6 \times \cancel{ {10}^{ - 9}} = \dfrac{ 500 \times \cancel{ {10}^{ - 9}} \times 1}{d}=>6×

10

−9

=

d

500×

10

−9

×1

= > d = \dfrac{500}{6}=>d=

6

500

= > d = 83.33 \: m=>d=83.33m

Distance of 5th bright fringe from central fringe will be:

\therefore \: D_{n} = \dfrac{n \lambda D}{d}∴D

n

=

d

nλD

= > \: D_{5} = \dfrac{5 \times 500 \times {10}^{ - 9} \times 1}{83.33}=>D

5

=

83.33

5×500×10

−9

×1

= > D_{5} = 30 \times {10}^{ - 9} m=>D

5

=30×10

−9

m

= > D_{5} = 30 \: nm=>D

5

=30nm

So final answer is :

d = 83.33 m

D_{5}D

5

= 30 nm