Q: What is the effect on the fringe width of interference fringes in a Young's double slit experiment due to each of the following operations.(a) The screen is moved away from the plane of the slits.(b) The (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength.(c) The separation between the two slits is increased(d) The width of two slits are slightly increased(In each operation, take all parameters, other than the one specified to remain unchanged)

Bandwidth;\$\beta= (\frac{D}{d}) \lambda\$ where D is Distance between screen to slits d is Distance between the slits\$\lambda\$ is wavelength(a) Screen is moved away--Here, D increases Hence, bandwidth increases(b) Wavelength decreases --here, \$\lambda\$decreases Hence, bandwidth decreases Wavelength increases -- Here, \$\lambda\$ increases Hence, bandwidth increases(c) Slit width increase -- Here, d increases Hence, bandwidth decreases (d) As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so widethen the interference pattern disappears.

Q: In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

\$d = 10^{-3}\:m\$\$D=1\:m\$\$ \lambda = 500 \times 10^{-9}\:m\$The width of central maxima in a single slit diffraction: \$\dfrac{2 \lambda D}{a}\$Fringe width in double slit pattern: \$\beta = \dfrac{\lambda D}{d}\$Given,\$10 \beta = \dfrac{2 \lambda D}{a}\$\$ \Rightarrow 10 \dfrac{\lambda D}{d} = \dfrac{2 \lambda D}{a}\$\$ \Rightarrow a =\dfrac{d}{5}\:mm= 0.2\:mm\$

Q: In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Distance between the slits, \$d=0.28\times10^{-3}\ m\$Distance between the slits and the screen, \$D = 1.4 m\$Distance between the central fringe and the fourth \$(n = 4)\$ fringe, \$u=1.2\times10^{-2}\ m\$In case of a constructive interference, we have the relation for the distance between the two fringes as:\$u=n\lambda D/d\$\$\Rightarrow \lambda=ud/nD=6\times10^{-7}m=600nm\$

Q: The ratio of amplitude of two waves is 4:3 then the ratio of maximum and minimum intensity is :

Let amplitudes \$4x\$ and \$3x\$ respectively.At maxima amplitudes add up giving \$7x\$At minima amplitudes substract giving \$x\$Intensities ratio \$=\dfrac{49}{1}\$Intensity is square of amplitude.

Question 1

What is the effect on the fringe width of interference fringes in a Young's double slit experiment due to each of the following operations.(a) The screen is moved away from the plane of the slits.(b) The (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength.(c) The separation between the two slits is increased(d) The width of  two slits are slightly increased(In each operation, take all parameters, other than the one specified to remain unchanged)

Accepted Answer

Bandwidth;$\beta= (\frac{D}{d}) \lambda$ where D is Distance between screen to slits d is Distance between the slits$\lambda$ is wavelength(a) Screen is moved away--Here, D increases  Hence, bandwidth increases(b) Wavelength decreases --here, $\lambda$decreases Hence, bandwidth decreases Wavelength increases -- Here, $\lambda$ increases Hence, bandwidth increases(c) Slit width increase -- Here, d increases Hence, bandwidth decreases (d) As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so widethen the interference pattern disappears.

Question 2

In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

Accepted Answer

$d = 10^{-3}\:m$$D=1\:m$$ \lambda = 500 \times 10^{-9}\:m$The width of central maxima in a single slit diffraction: $\dfrac{2 \lambda D}{a}$Fringe width in double slit pattern: $\beta = \dfrac{\lambda D}{d}$Given,$10 \beta = \dfrac{2 \lambda D}{a}$$ \Rightarrow 10 \dfrac{\lambda D}{d} =  \dfrac{2 \lambda D}{a}$$ \Rightarrow a =\dfrac{d}{5}\:mm= 0.2\:mm$

Question 3

Derive the expression for the fringe width in a Young's double slit experiment. How will the fringe width change if (i) separation between the slits is increased (ii) screen is moved away from the plane of the slits.

Accepted Answer

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Question 4

Using Huygen's wave theory, derive Snell's law of refraction.

Accepted Answer

Snell's law of refraction using Huygen's Principle:Consider a plane wavefront $AB$ incident on a surface $PQ$ separating two media 1 and 2.The medium 1 is a rarer medium of refractive index $n_1$ in which light travels with a velocity $C_1$. The medium 2 is denser medium of refractive index $n_2$ in which light travels with a velocity $C_2$.The angle between the incident ray $FA$ and the normal $NA$ at the point of incidence $A$ is equal to $i$.The angle is also equal to the angle between the incident plane wavefront $BA$ and the surface of separation $PQ$.So $\angle BAD$ is the angle of incidence of the incident plane wavefront $AB$.Similarly, the angle between the refracted wavefront and the surface of separation $PQ$ is equal to the angle of refraction $r$.i.e. $\angle ADC = r$.Consider the triangles $BAD$ $ACD$ figure above.$\sin  i = \sin \angle BAD = \dfrac{BD}{AD} = \dfrac{C_1r}{AD}$$\sin r = \sin \angle ADC = \dfrac{AC}{AD}= \dfrac{C_2r}{AD}$$\dfrac{\sin i}{\sin r} = \dfrac{C_1t}{C_2t} = \dfrac{C_1}{C_2}=$ constantThis constant is called the refractive index of the second medium (2) with respect to the first medium (1).$\dfrac{C_1}{C_2} = \dfrac{n_2}{n_1} = 1n_2$This equation proves the Snell's law.

Question 5

In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Accepted Answer

Distance between the slits, $d=0.28\times10^{-3}\ m$Distance between the slits and the screen, $D = 1.4 m$Distance between the central fringe and the fourth $(n = 4)$ fringe, $u=1.2\times10^{-2}\ m$In case of a constructive interference, we have the relation for the distance between the&#10;two fringes as:$u=n\lambda D/d$$\Rightarrow \lambda=ud/nD=6\times10^{-7}m=600nm$

Question 6

What is the shape of the wavefront in each of the following cases:(a) Light diverging from a point source.(b) Light emerging out of a convex lens when a point source is placed at its focus.(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Accepted Answer

(a) Spherical, The point source emits the spherical wavefront as the light diverges from a point source in all directions.(b) Planar, The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.(c) Planar, (A small area on the surface of a large sphere is nearly planar). The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Question 7

What is the effect on the interference fringes in a Young's double slit experiment when (i) the width of the two slits is increased (ii) the monochromatic source is replaced by a source of white light and (iii) the separation between the two slits is increased, keeping other variables constant in each case?

Accepted Answer

(i) : Intensity of light emitted by the slits increases as the width of the slit is increased. Thus, more brighter fringes are formed at the screen on increasing the width of the slits.(ii) : Coloured fringes are formed at the screen if monochromatic light is replaced by the white light.(iii) : Fringe width   $\beta = \dfrac{\lambda \ D}{d}$So, width of the fringes gets decreased if the distance between the slits $(d)$ is increased and thus we get narrower fringes.

Question 8

Draw the diagrams to show the behaviour of plane wave fronts as they (a) pass through a thin prism, and (b) pass through a thin convex lens and (c) reflect by a concave mirror.

Accepted Answer

The behavior of a thin prism and a thin convex lens are shown in figs. (a) and (b) respectively.

Question 9

The ratio of amplitude of two waves is 4:3 then the ratio of maximum and minimum intensity is :

Accepted Answer

Let amplitudes $4x$ and $3x$ respectively.At maxima amplitudes add up giving $7x$At minima amplitudes substract giving $x$Intensities ratio $=\dfrac{49}{1}$Intensity is square of amplitude.

Question 10

Which of the following phenomenon works on the particle nature of light?

Accepted Answer

(a) Interference is a phenomenon in which two waves of same frequency superpose to give resultant intensity different from sum of their separate intensity. So, it cannot exhibit particle's nature of light.(b) Diffraction is a phenomenon in which light bends at sharp ends of an obstacle or a hole. So, it also cannot exhibit particle's nature of light.(c) Polarisation of light is a property owing to which a light ray after emerging through a crystal (a special kind like tourmaline) have vibrations in a plane perpendicular to its direction of propagation. So, it also can not explain particle's nature of light.(d) Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particles's nature of lighthence (d) is correct.