Question

Light of wavelength 5500 from a narrow slit is incident on a double slit. The overall separation of 5 fringes on a screen 200 cm away is 1 cm, calculate (a) the slit separation and (b) the fringe width

Answer 1

Answer:

The distance between the slit is 5.5 × $10^{-4}$ m and Fringe width is 1 × $10^{-3}$ m

Explanation:

Given: Light of wavelength is  5500

Distance between 5 fringes on a screen 1 cm away from 200 cm

To find: Distance between the slit and Fringe width

Solution:

Wavelength of light:

The distance between corresponding spots in two successive light cycles is known as the wavelength of light.

Given that ,

Wavelength λ =  5500 × $10^{-10}$ m

10 β  = 1 cm

β = 0.01 m

 = 1 × $10^{-3}$

D = 200 cm

β  = λD /d

1 × $10^{-3}$ = $\frac{(5500 * 10^{-10} )(2) }{d}$

d = $\frac{(5500 *10^{-10} )(2)}{1 * 10^{-3} }$

d =  $5.5 * 10 ^{-4}$ m

The distance between the slit is 5.5 × $10^{-4}$ m

Final answer:

The distance between the slit is 5.5 × $10^{-4}$ m and Fringe width is 1 × $10^{-3}$ m

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Answer 2

Answer:

a) Slit separation = 0.55 x $10^{-3}$ m ; b) fringe width β = 2 x $10^{-3}$ m

Explanation:

In Young's Double slit experiment, the given values are as follows:

λ = 5500 x $10^{-10}$m

n = 5

D = 200 cm = 2m [Distance between the screen and slit]

from the given data it is clear that,

The separation between 5 fringes is given as 1cm

5β = 1

a) β = $\frac{1}{5}$ = 0.2 cm = 2 x $10^{-3}$ m

Thus the fringe width β is  2 x $10^{-3}$ m

To find the slit width,

β = λD/d

Thus, d = λD/β

On substituting the values of λ, D and β we get

b) d = (5500 x $10^{-10}$ x 2) / 2 x $10^{-3}$ m

  =  5500 x $10^{-7}$ m

  =  0.55 x $10^{-3}$m

Therefore, the slit separation is 0.55 x $10^{-3}$m

On the whole, the slit separation is 0.55 x $10^{-3}$ m and fringe width β is 2 x $10^{-3}$ m.

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