Question

White coherent light (400 nm-700 nm) is sent through the slits of a Young’s double slit experiment (figure 17-E3). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?

Answer 1

Given:

Separation between two slits=d=0.5 mm=0.5×10⁻³ m

Wavelength of the light λ=400 nm to 700 nm

Distance of the screen from the slit D=50 cm=0.5 m,

Position of hole on the screen,

yn=1 mm=1×10⁻³ m

a)we know that for zero intensity ( dark fringe )

yn= [(2n+1)  /2λn ]x D/d where n=012....

λn= [2/2n+1 ] x ynd/D

= 2/ (2n+1) x 10⁻³ x(0.05) x 10⁻³ / 0.5

=2/ (2n+1) x 10 ⁻⁶m

=[2/ (2n+1)]  x 10 ³ nm

if n=1 λ1= (2/3) x 1000=667nm

if n=1  λ2= (2/5)  x 1000= 400nm

Thus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole.

b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.

So, yn=nλnD/d

λn=ynd/nD

when n=1

λ1= yn d/D =10 ⁻³ x(0.5) x 10 ⁻³/0.5

10⁻³m=100nm

1000nm is ot resent in the range 400nm - 700nm

again  when n=2 λ2= yn d/2D=500nm

so wavelength which will have strong intensity is 500nm

Answer 2

(a) Wavelength(s) which will be absent in the light coming from the hole are $400 \mathrm{nm} \text { and } 667 \mathrm{nm}$

(b) Wavelength(s) which will have a strong intensity is $500 \mathrm{nm}$

Explanation:

Divide between 2 slits $=d=0.5 \mathrm{mm}=0.5 \times 10^{-3} \mathrm{m}$

$\lambda=400 \mathrm{nm} \text { to } 700 \mathrm{nm}$

λ  is The Light wavelength

$D=50 \mathrm{cm}=0.5 \mathrm{m}$

D is Screen gap from the slit  

Hole location on screen,

$y n=1 m m=1 \times 10^{-3} m$

(a) We know this for intensity zero ( dark fringe )

$\mathrm{yn}=\left[\frac{(2 \mathrm{n}+1)}{2 \lambda \mathrm{n}}\right] \times \frac{\mathrm{D}}{\mathrm{d}} \text { where } \mathrm{n}=012 \ldots .$

$\lambda n=\left[\frac{2}{2 n+1}\right] \times \frac{\text { ynd }}{D}$

$=\left[\frac{2}{2 n+1}\right] \times 10^{-3} \times(0.05) \times \frac{10^{-3}}{0.5}$

$=\left[\frac{2}{2 n+1}\right] \times 10^{-6} m$

$=\left[\frac{2}{2 n+1}\right] \times 10^{3} n m$

$\text { if } \mathrm{n}=1, \lambda 1=\left(\frac{2}{3}\right) \times 1000=667 \mathrm{nm}$

$\text { if } \mathrm{n}=1, \lambda 2=\left(\frac{2}{3}\right) \times 1000=400 \mathrm{nm}$

Therefore the light waves between $400 \mathrm{nm}$ and $667 \mathrm{nm}$ of wavelength will be absent from the light coming from the cavity.

(b) The wavelength(s) will be of a high intensity, creating a bright fringe at the hole spot.

$\text { So, yn }=\frac{n \lambda n D}{d}$

$\lambda \mathrm{n}=\frac{\mathrm{ynd}}{\mathrm{n} \mathrm{D}}$

when $n=1$

$\lambda 1$$=\mathrm{yn}$$\frac{d}{D}=10^{-3}$$\times(0.5) \times \frac{10^{-3}}{0.5}$

$10^{-3} \mathrm{m}=100 \mathrm{nm}$

$1000 \mathrm{nm}$ is within $400 \mathrm{nm}-700 \mathrm{nm}$ range

again  when $n=2 \lambda 2=$$\text { yn } \frac{\mathrm{d}}{2 \mathrm{D}}=500 \mathrm{nm}$

So the wavelength of $500 \mathrm{nm}$ that will have a high intensity