## In young’s double slit experiment one of the slits is wider than other, so that amplitude of the light from one slit is…..

Q. In young’s double slit experiment one of the slits is wider than other, so that amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I, when they interfere at phase difference φ is given by

(a) Im/9(4 + 5 cosφ )

(b) Im/3 (1+2 cos2 (φ/2 ))

(c) Im/5 (1+4 cos2 (φ/2))

(d) Im/9 (1+8 cos2 (φ/2 ))

Ans: (d)

Sol: a1 = 2 a2

I1 = 4I’ & I2 = I’

$\displaystyle I_m = (\sqrt {I_1} + \sqrt {I_2} )^2$

$\displaystyle I_m = (\sqrt {4I’} + \sqrt {I’} )^2 = 9I’$

$\displaystyle I’ = \frac{I_m}{9}$ ….(i)

$\displaystyle I = I_1 + I_2 + 2\sqrt{I_1 I_2} cos\phi$

$\displaystyle I = 4I’ + I’ + 2\sqrt{4I’ \times I’} cos\phi$

$\displaystyle = 5I’ + 4I’ cos\phi$

$\displaystyle = 5I’ + 4I'(2 cos^2 \phi/2 – 1 )$

$\displaystyle = I’ + 8I’cos^2 \phi/2$

$\displaystyle = I'(1 + 8 cos^2 \phi/2 )$

From(i)

$\displaystyle I = \frac{I_m}{9}(1 + 8 cos^2 \phi/2 )$

## A beam of unpolarized light of intensity I_0 is passed through a polaroid A and then through another polaroid

Q: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that it principal plane makes an angle of 45° relative to that of A . The intensity of emergent light is

(a) I0

(b) I0/2

(c) I0/4

(d) I0/8

Ans: (c)

Solution: Here, polaroid A polarizes light.

Intensity of polarized light from A= I0/2

According to law of Malus ,

Intensity of light emerging from B,

$\displaystyle I_B = \frac{I_0}{2}cos^2 45^o$

$\displaystyle I_B = \frac{I_0}{2}(\frac{1}{\sqrt{2}})^2$

$\displaystyle I_B = \frac{I_0}{4}$

## Two waves are represented by :  y1 = a1 cos (ωt – kx) and y2 = a2 sin(ωt – kx + π/3)….

Q: Two waves are represented by :

y1 = a1 cos (ωt – kx) and y2 = a2 sin(ωt – kx + π/3)

Then the phase difference between them is :

Solution : y1 = a1 cos (ωt – kx)

y1 = a1 sin (ωt – kx + π/2 )

and y2 = a2 sin(ωt – kx + π/3)

Phase difference $\large \phi = \frac{\pi}{2} – \frac{\pi}{3}= \frac{\pi}{6}$

## Disturbances of two waves are shown as a function of time in the following figure. The ratio of their intensities will be

Q: Disturbances of two waves are shown as a function of time in the following figure. The ratio of their intensities will be –

Solution : $\large I \propto f^2 A^2$

$\large \frac{I_1}{I_2} = (\frac{f_1}{f_2})^2 (\frac{A_1}{A_2})^2$

$\large \frac{I_1}{I_2} = (\frac{2f}{f})^2 (\frac{a}{2a})^2$

$\large \frac{I_1}{I_2} = 1 : 1$

## Two coherent sources separated by distance d are radiating in phase having wavelength λ and are placed on x-axis…

Q: Two coherent sources separated by distance d are radiating in phase having wavelength λ and are placed on x-axis
symmetrically about the origin A detector moves in a big circle (centre origin) around the two sources in the plane of the two sources. The angular position of n = 4 interference maxima is given as

Solution :

Here path difference at a point P on the circle is given by

Δx = d cosθ …(i)

For maxima at P

Δx = n λ …(ii)

From equation (i) and (ii)

d cosθ = n λ

$\large cos\theta = \frac{n \lambda}{d}$

$\large \theta = cos^{-1} (\frac{n \lambda}{d})$

## Light of wavelength λ in air enters a medium of refractive index μ. Two points in this medium, lying along the path of this light ….

Q: Light of wavelength λ in air enters a medium of refractive index μ . Two points in this medium, lying along the path of this light are at a distance x apart. The phase difference between these points is :

Solution :

In Medium Path length length becomes Δx = μ x

Phase difference $\large \phi = \frac{2 \pi}{\lambda} \times \Delta x$

$\large \phi = \frac{2 \pi}{\lambda} \times \mu x$

## A monochromatic light of λ = 5000 A° is incident on two slits separated by a distance of 5 × 10^–4 m. The interference pattern in seen on a screen …

Q: A monochromatic light of λ = 5000 A° is incident on two slits separated by a distance of 5 × 10–4 m. The interference pattern in seen on a screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m & refractive index μ = 1.5 is placed between one of the slits & the screen. Find the intensity at the centre of the screen, if the intensity there is I0 in the absence of the plate. Also find the lateral shift of the central maximum.

Solution : When a thin glass plate of thickness t & refractive index μ is introduced between one of slit & screen ;

Path difference $\large \Delta x = (\mu – 1) t$

$\large \Delta x = (1.5 – 1) 1.5 \times 10^{-6}$

$\large \Delta x = 0.75 \times 10^{-6}$

Phase difference , $\large \phi = \frac{2\pi}{\lambda}\times \Delta x$

$\large \phi = \frac{2\pi}{5 \times 10^{-7}}\times 0.75 \times 10^{-6}$

φ = 3 π

$\large I = I_0 cos^2 \frac{\phi}{2}$

$\large I = I_0 cos^2 \frac{3 \pi}{2} = 0$

$\large Shift = \frac{D}{d}(\mu -1)t$

$\large y_o = \frac{D}{d}\Delta x$

$\large y_o = \frac{1}{5 \times 10^{-4}}\times 0.75 \times 10^{-6}$

$\large y_o = 1.5 \times 10^{-3} m$

## A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm & distance between the plane of the slits & screen is 1.33 m…

Q: A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm & distance between the plane of the slits & screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 Å.
(a) Calculate the fringe width.
(b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis.

Solution :

$\large \frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$

$\large \frac{1.33}{1} = \frac{6300}{\lambda_2}$

$\large \lambda_2 = \frac{6300}{1.33}$

$\large \beta = \frac{\lambda_2 D}{d}$

$\large \beta = \frac{6300 \times 1.33}{1.33 \times 10^{-3}}$

= 0.63 mm

(b) $\large \Delta x = (\frac{\mu’}{\mu}-1)t$

$\large \frac{\lambda}{2} = (\frac{1.53}{1.33}-1)t$

$\large \frac{6300}{1.33 \times 2} = (\frac{1.53}{1.33}-1)t$

t = 1.575 μm

## A young’s double slit experiment is performed using light of wavelength λ = 5000 A° . which emerges in phase from two slits …

Q: A young’s double slit experiment is performed using light of wavelength λ = 5000 A° . which emerges in phase from two slits a distance d = 3 × 10–7 m apart. A transparent sheet of thickness t = 1.5 × 10–7 m is placed over one of the slits. The refractive index of the material of this sheet is μ = 1.17. Where does the central maximum of the interference pattern now appear ?

Solution : Path difference Δx = (μ – 1)t

= (1.17–1)(1.5 × 10–7) = 0.255 × 10–7

Now for central maxima :

$\large \Delta x = \frac{y d}{D}$

$\large 0.255 \times 10^{-7} = \frac{y \times 3 \times 10^{-7}}{D}$

y = 0.085 D

y/D = 0.085

Alternate sol:

$\large \Delta x = d sin\theta$

$\large Sin\theta = \frac{\Delta x}{d}$

$\large Sin\theta = \frac{0.255 \times 10^{-7}}{3 \times 10^{-7}}$

$\large Sin\theta = \frac{0.255 }{3 } rad$

$\large Sin\theta = 0.085 rad$

θ = 4.88°

## In a bi-prism experiment with sodium light, bands of width of 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens ….

Q: In a bi-prism experiment with sodium light, bands of width of 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens 30 cm away from the slit between and screen, two images of the slit are seen 0.7 cm apart at 100 cm distance from the slit. Calculate the wavelength of sodium light.

Solution : When convex lens is introduced ,

v = 70 cm , u = – 30 cm

$\large |\frac{v}{u}| = |\frac{h_i}{h_o}|$

$\large \frac{+7}{3} = \frac{0.7 cm}{h_o}$

h0 = 0.3 cm i.e. d = 0.3 cm

$\large \frac{\lambda D}{d} = 0.0195$

$\large \frac{\lambda (100)}{0.3} = 0.0195$

λ = 5850 A°