Question

(a) In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.

(b)Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.

(c) What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm?

OR

(a) Two thin convex lenses $L_1$and $L_2$ of focal lengths $f_1$and$f_2$ respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens $L_1$ . Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system.

(b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such that AQ = AR

Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is ${60}^0$ and refractive index of the material of prism is $\sqrt{3}$. Determine the angle of incidence and angle of deviation

Answer 1

(a) Young’s double slit experiment demonstrated the phenomenon of interference of light. Consider two fine slits $S_1$and $S_2$ at a small distance d apart. Let the slits be illuminated by a monochromatic source of light of wavelength $\lambda$.Let GG′ be a screen kept at a distance D from the slits. The two waves emanating from slits $S_1$and $S_2$ superimpose on each other resulting in the formation of an interference pattern on the screen placed parallel to the slits.Let O be the centre of the distance between the slits. The intensity of light at a point on the screen will depend on the path difference between the two waves reaching that point. Consider an arbitrary point P at a distance x from O on the screen.Path difference between two waves at $P=S_{2}P-S_{1}P$

The intensity at the point P is maximum or minimum as the path difference is an integral multiple of wavelength or an odd integral multiple of half wavelength. For the point P to correspond to maxima, we must have

$S_{2}P-S_{1}P=n\lambda ,n=0,1,2,\mathrm{3....}$

From the figure given above,

$(S_{2}P{)}^2-(S_{1}P{)}^2=D^2+{(x+\frac{d}{2})}^2-D^2-{(x-\frac{d}{2})}^2$

On solving we get

$(S_{2}P{)}^2-(S_{1}P{)}^2=2xd{S}_{2}P-S_{1}P=\frac{2xd}{S_{2}P+S_{1}P}Asd<<D,then{S}_{2}P+S_{1}P=2D(\because S_{1}P=S_{2}P\cong D)\therefore S_{2}P-S_1=\frac{2xd}{2d}=\frac{xd}{D}Pathdifference{S}_2-S_{1}P=\frac{xd}{D}$

Hence, when constructive interference occur, bright region is formed.For maxima or bright fringe, path difference $\frac{xd}{D}=n\lambda$

$i.ex=\frac{n\lambda D}{d}$

where n=0, ±1, ±2,........

During destructive interference dark fringes are formed:

Path difference $\frac{xd}{D}=(n+\frac{1}{2})\lambda$

$x=(n+\frac{1}{2})\frac{\lambda D}{d}$

The dark fringe and the bright fringe are equally spaced and the distance between consecutive bright and dark fringe is given by:

$\beta =x_{n+1}-x_n\beta =\frac{(n+1)\lambda D}{d}-\frac{n\lambda D}{d}\beta =\frac{\lambda D}{d}$

Hence the fringe width is given by $\beta =\frac{\lambda D}{d}$

(b) The intensity variation in the fringe pattern obtained on a screen in a Young’s double slit experiment corresponds to both single slit diffraction and double slit interference because the two sources are slits of finite width in the double slit

experiment. If a lens is placed in front of the double slits and when one of the slit $S_1$ is closed and the other kept open, a single slit diffraction is formed on the screen. A similar diffraction pattern is obtained on the screen if the slit $S_1$ is kept open and $S_2$ is closed. Both diffraction patterns form on the same position on the screen in the focal plane of the lens. When both slits open simultaneously, the resulting total intensity pattern on the screen is actually the superposition of the single slit diffraction pattern formed by waves from various point sources of each slit and a double slit interference pattern as shown. The actual double slit intensity pattern consists of the interference pattern (solid lines) formed within the diffraction pattern (dotted lines).

(c) Let the width of each slit be ‘a’.

The separation between m maxima in a double slit experiment is given by ym $y_m=m\frac{\lambda D}{d}$

where D is the distance between the screen and the slit and d is the separation between the slits. We know that the angular separation between m maxima can be given as

${\theta }_m=\frac{y_m}{D}=\frac{m\frac{\lambda D}{d}}{D}$

$\Rightarrow {\theta }_m=\frac{m\lambda }{d}$

Therefore, we can write the angular separation between 10 bright fringes as

$\Rightarrow {\theta }_{10}=\frac{10\lambda }{d}----------\left(1\right)$

The angular width of the central maximum in the diffraction pattern due to a single slit of width ‘a’ is given by

$2{\theta }_1=2\left(\frac{\lambda }{a}\right)----------\left(2\right)$

It is given that 10 maxima of the double slit pattern is formed within the central maximum of the single slit pattern.Therefore, we can equate (1) and (2) as

$\frac{10\lambda }{d}=\frac{2\lambda }{a}$

Solving, we get

$a=\frac{d}{5}$

Given that the seperation between the slits = 1 mm

Therefore the slit width $a=\frac{d}{5}$ = $\frac{1}{5}=0.2mm$

Therefore, the width of each slit = 0.2 mm

OR

(a) Consider two thin lens $L_1$ and $L_2$ of focal length $f_1$ and $f_2$ held coaxially in contact

with each other. Let P be the point where the optical centres of the lenses coincide (lenses being thin).Let the object be placed at a point O beyond the focus of lens $L_1$ such that OP = u(object distance). Lens $L_1$ alone forms the image at $I_1$ where P $I_1$ = $v_1$ (image distance). The image $I_1$ would serve as a virtual object for lens $L_2$ which forms a final image I at distance P I = v. The ray diagram showing the image formation by the combination of these two thin convex lenses will be as shown below:

From the lens formula, for the image $I_1$ formed by the lens $L_1$, we have

$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}.............\left(1\right)$

for the image formation by the second lens $L_2$

$\frac{1}{v}-\frac{1}{V_1}=\frac{1}{f_2}.............\left(2\right)$

Adding (1) and (2) we get :

$\frac{1}{v_1}+\frac{1}{u}=\frac{1}{v}+\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{1}{u}+\frac{1}{v}=\frac{1}{f_1}+\frac{1}{f_2}\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$

If the two lenses are considered a single lens of focal length f, which forms an image I at a distance v with an object distance being u, then we get

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}(where,\frac{1}{f}=\frac{1}{f_1}=\frac{1}{f})\Rightarrow f=\frac{f_1{f}_2}{f_1+f_2}$

Hence, the focal length of the combined system is given by

f=$\Rightarrow f=\frac{f_1{f}_2}{f_1+f_2}$

(b) Given that side AQ = AR. This implies that $\angle$AQR = $\angle$ARQ.The ray diagram for the refraction of ray PQ passing through the prism ABC is as shown below.

As the ray PQ after refraction from surface AB emerges from face AC at point R of the prism, it implies that the refracted ray QR travels parallel to the base of the prism. This happens at the minimum deviation position.So, according to the angle of minimum deviation formula, we have

$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\frac{A}{2}}---------\left(1\right)$

Where A is the angle of prism,${\delta }_m$is the angel of minimum deviation and$\mu$ is the refractive index of the prism.

Given $A={60}^{\circ },n=\sqrt{3}$

Substituting in (1) we get:

$\sqrt{3}=\frac{sin\frac{({60}^{\circ }+{\delta }_m)}{2}}{sin\frac{{60}^{\circ }}{2}}\sqrt{3}\times {30}^{\circ }=sin\frac{({60}^{\circ }+{\delta }_m)}{2}\frac{\sqrt{3}}{2}=sin\frac{({60}^{\circ }+{\delta }_m)}{2}si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{({60}^{\circ }+{\delta }_m)}{2}{60}^{\circ }=\frac{({60}^{\circ }+{\delta }_m)}{2}{120}^{\circ }=({60}^{\circ }+{\lambda }_m)\Rightarrow {\delta }_m={60}^{\circ }{60}^{\circ }$

Thus, the angle of minimum deviation=${60}^{\circ }$ .At the minimum deviation position

$i=\frac{A+{\delta }_m}{2}A={60}^{\circ },{\delta }_m={60}^{\circ }i=\frac{{60}^{\circ }+{60}^{\circ }}{2}\Rightarrow i={60}^{\circ }$