Q: In Youngs double-slit experiment using monochromatic light of wavelength \$\lambda\$, the intensity of light at a point on the screen where path difference is \$\lambda\$, is K units. What is the intensity of light at a point where path difference is \$\lambda\$/3?

For monochromatic light, \$I_1+I_2\$ is the intensity.\$I^{'}=I_1+I_2+2\sqrt{I_1I_2}cos\theta\$Phase difference is given by:\$\phi=2\pi\times \dfrac{path\ difference}{\lambda}\$\$\phi=2\pi\times\dfrac{\lambda}{\lambda}\$\$=2\pi\$So, the new intensity can be obtained as: \$\Rightarrow I^{'}=4I_1\$given \$I^{'}=K,I_1=K/4,path\ difference=\lambda/3\$phase difference is \$2\pi\times 1/3\$Hence, \$I=I_1+I_2+2\sqrt{I_1I_2}cos2\pi/3\$\$\Rightarrow I=I_1=K/4\$

Q: A beam of light of wavelength \$600\ nm\$ from a distance source falls on a single slit \$1\ mm\$ wide and a resulting diffraction pattern is observed on a screen \$2\ m\$ away. The distance between the first dark fringes on either side of central bright fringe is

for a dark fringe to form\$\dfrac{dy}{D}=\lambda\$\$y=\dfrac{D\lambda}{d}=\dfrac{2\times 600 \times 10^{-9}}{10^{-3}}=1.2mm\$distance between the first dark fringes on either side of central bright fringe is \$2\times y=2.4mm\$option \$D\$ is correct

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: In Young's double slit experiment, the intensity of light at a point on the screen where the path difference '\$\lambda \$' is 'K' units. The intensity of light at a point where the path difference is \$\dfrac{\lambda}{3} \$ is (\$\lambda \$ being the wavelength of light used)

phase difference corresponding to path difference of \$\lambda /3\$ is\$\phi =\dfrac{2\pi }{3}\$So, \$I=k cos^{2}(\dfrac{2\pi /3}{2}) (\because I=I_{0} cos^{2}(\phi /2))\$ \$=k(+1/2)^{2} (\because cos \pi /3=1/2)\$ \$=\dfrac{k}{4}\$

Question 1

In Youngs double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is K units. What is the intensity of light at a point where path difference is $\lambda$/3?

Accepted Answer

For monochromatic light, $I_1+I_2$ is the intensity.$I^{'}=I_1+I_2+2\sqrt{I_1I_2}cos\theta$Phase difference is given by:$\phi=2\pi\times \dfrac{path\ difference}{\lambda}$$\phi=2\pi\times\dfrac{\lambda}{\lambda}$$=2\pi$So, the new intensity can be obtained as: $\Rightarrow I^{'}=4I_1$given $I^{'}=K,I_1=K/4,path\ difference=\lambda/3$phase difference is $2\pi\times 1/3$Hence, $I=I_1+I_2+2\sqrt{I_1I_2}cos2\pi/3$$\Rightarrow I=I_1=K/4$

Question 2

A beam of light of wavelength $600\ nm$ from a distance source falls on a single slit $1\ mm$ wide and a resulting diffraction pattern is observed on a screen $2\ m$ away. The distance between the first dark fringes on either side of central bright fringe is

Accepted Answer

for a dark fringe to form$\dfrac{dy}{D}=\lambda$$y=\dfrac{D\lambda}{d}=\dfrac{2\times 600 \times 10^{-9}}{10^{-3}}=1.2mm$distance between the first dark fringes on either side of central bright fringe is $2\times y=2.4mm$option $D$ is correct

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

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference '$\lambda $' is 'K' units. The intensity of light at a point where the path difference is $\dfrac{\lambda}{3} $ is ($\lambda $ being the wavelength of light used)

Accepted Answer

phase difference corresponding to path difference of $\lambda /3$ is$\phi =\dfrac{2\pi }{3}$So, $I=k cos^{2}(\dfrac{2\pi /3}{2})     (\because I=I_{0} cos^{2}(\phi /2))$         $=k(+1/2)^{2}     (\because cos \pi /3=1/2)$         $=\dfrac{k}{4}$

Question 9

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 10

Answer the following questions:(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.(e) Ray optics is based on the assumption that light travels in a straight line.  Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

Accepted Answer

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.