Wave Optics Concept Page - 6

Example

Define and calculate fringe width and angular fringe width in YDSE

Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. It is given by:
β=λDd$\beta =\frac{\lambda D}{d}$
Angular fringe width is given by:
tanθ≈θ=βD=λd$\tan \theta \approx \theta =\frac{\beta }{D}=\frac{\lambda }{d}$
Example: In Young's double slit experiment the two slits are illuminated by light of wavelength 5890∘A${5890}^{\circ }A$ and the distance between the fringes obtained on the screen is 0.2∘${0.2}^{\circ }$. If the whole apparatus is immersed in water then find the angular fringe width. (refractive index of water is 4/3$4/3$)

Solution:
For interference in YDSE:
dsinθ=nλ$d\sin \theta =n\lambda$
For small θ$\theta$, sinθ$\sin \theta$ can be approximated to θ$\theta$
So, dθ=nλ$d\theta =n\lambda$
or, θ∝λ$\theta \propto \lambda$
Now as the set up immersed in water, λ$\lambda$ will become λμ=λ4/3=3λ4${\displaystyle \frac{\lambda }{\mu }}={\displaystyle \frac{\lambda }{4/3}}={\displaystyle \frac{3\lambda }{4}}$
So, λ1λ2=θ1θ2${\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}={\displaystyle \frac{{\theta }_1}{{\theta }_2}}$
or, λ3λ/4=0.2∘θ2${\displaystyle \frac{\lambda }{3\lambda /4}}={\displaystyle \frac{{0.2}^{\circ }}{{\theta }_2}}$
or, θ2=0.2∘×34=0.15∘.${\theta }_2={0.2}^{\circ }\times {\displaystyle \frac{3}{4}}={0.15}^{\circ }.$

Example

Solve problems on interference on a screen where initial path difference is given

Example: In Youngs double slit interference experiment the wavelength of light used is 6000A∘$6000A^{\circ }$. If the path difference between waves reaching a point P on the screen is 1.5 micron$1.5micron$, then at that point P which dark band will occur?

Solution:
Here Δxλ${\displaystyle \frac{\mathrm{\Delta }x}{\lambda }}$ =1.5×10−66000×10−10$={\displaystyle \frac{1.5\times {10}^{-6}}{6000\times {10}^{-10}}}$=2.5$=2.5$=52$={\displaystyle \frac{5}{2}}$
or Δx=52λ$\mathrm{\Delta }x={\displaystyle \frac{5}{2}}\lambda$
So, it corresponds to third minima or third dark band because path difference λ2${\displaystyle \frac{\lambda }{2}}$ corresponds to first dark band, 3λ2${\displaystyle \frac{3\lambda }{2}}$ corresponds to second dark band, 5λ2${\displaystyle \frac{5\lambda }{2}}$ corresponds to third dark band and so on.

Example

Solve problems on interference on a screen where two coherent sources are created by reflection

Example: In figure, PQ represents a plane wavefront and AO and BP the corresponding extreme rays of monochromatic light of wavelength λ$\lambda$. Find the value of angle θ$\theta$ for which the ray BP and the reflected ray OP interfere constructively.

Solution:
Since, P$P$ and Q$Q$ are points on the same wavefront, they are in the same phase. Therefore, the path difference at point P between the ray BP and the reflected ray OP is δx=QO+OP$\delta x=QO+OP$
Now, in triangle POR, OP=PRcosθ=dcosθ$OP={\displaystyle \frac{PR}{\cos \theta }}={\displaystyle \frac{d}{\cos \theta }}$
Also, in triangle QOP, QO=OPsin(90−2θ)=OPcos2θ$QO=OP\sin (90-2\theta )=OP\cos 2\theta$
∴δx=OPcos2θ+OP=OP(cos2θ−1)=2OPcos2θ=2×dcosθ×cos2θ=2dcosθ$\therefore \delta x=OP\cos 2\theta +OP=OP(\cos 2\theta -1)=2OP{\cos }^2\theta =2\times {\displaystyle \frac{d}{\cos \theta }}\times {\cos }^2\theta =2d\cos \theta$
Since there is a sudden path change of λ2${\displaystyle \frac{\lambda }{2}}$ due to reflection, the condition for constructive interference at P is
δx=λ2,3λ2$\delta x={\displaystyle \frac{\lambda }{2}},{\displaystyle \frac{3\lambda }{2}}$ etc.
Or 2dcosθ=λ2,3λ2$2d\cos \theta ={\displaystyle \frac{\lambda }{2}},{\displaystyle \frac{3\lambda }{2}}$ etc.
Or cosθ=λ4d,3λ4d$\cos \theta ={\displaystyle \frac{\lambda }{4d}},{\displaystyle \frac{3\lambda }{4d}}$ etc.
For constructively interference at P of BP and of ray path difference dcosθ=3λ4$dcos\theta ={\displaystyle \frac{3\lambda }{4}}$
or cosθ=3λ4d.$cos\theta ={\displaystyle \frac{3\lambda }{4d}}.$

Example

Problems related to change of intensity because of change of slit width

Example: In a Young's double-slit experiment, let β$\beta$ be the fringe width and let I0$I_0$ be the intensity at the central bright fringe. At a distance x$x$ from the central bright fringe, then find the intensity.

Solution:
△$\mathrm{△}$ = xdD$x{\displaystyle \frac{d}{D}}$, where △$\mathrm{△}$ is path difference between two waves.
∴${\displaystyle \therefore }$ phase difference = ϕ$\varphi$ = 2πλ△${\displaystyle \frac{2\pi }{\lambda }\mathrm{△}}$.
Let a=$a=$ amplitude at the screen due to each slit.
∴${\displaystyle \therefore }$ I0=k(2a)2$I_0=k(2a{)}^2$ =4ka2$=4k{a}^2$, where k$k$ is a constant.
For phase difference ϕ$\varphi$, amplitude =A=2acos(ϕ2$=A=2a\cos ({\displaystyle \frac{\varphi }{2}}$).
[Since, a2=a21+a22+2a1a2cosπ$a^2=a_1^2+a_2^2+2a_1{a}_{2}cos\pi$, here a1=a2$a_1=a_2$]
Intensity: I=KA2=k(4a2)cos2(ϕ2)=I0cos2(πλ.xdD)=I0cos2(πxβ)$I=K{A}^2=k(4a^2){\cos }^2({\displaystyle \frac{\varphi }{2}})=I_0{\cos }^2({\displaystyle \frac{\pi }{\lambda }}.{\displaystyle \frac{xd}{D}})=I_{0}co{s}^2({\displaystyle \frac{\pi x}{\beta }})$

Example

Problems on interference in glass slab

Example: A ray of light of intensity I$I$ incidents on a parallel glass slab at a point A, as shown in the given figure. It undergoes partial reflection and refraction. At each reflection, 25%$25\mathrm{\%}$ of incident energy is reflected. The rays AB and A' B' undergo interference. Then find the ratio ImaxImin${\displaystyle \frac{I_{max}}{I_{min}}}$ .

Solution:
Intensity of AB is 25%$25\mathrm{\%}$ of I1=I×25100=I4${\displaystyle I_1=I\times {\displaystyle \frac{25}{100}}=\frac{I}{4}}$
Intensity of A'B' is 75% of I1×25%×75%${\displaystyle I_1\times 25\mathrm{\%}\times 75\mathrm{\%}}$
=I×75100×25100×75100${\displaystyle =I\times \frac{75}{100}\times \frac{25}{100}\times \frac{75}{100}}$
=I×34×14×34${\displaystyle =I\times \frac{3}{4}\times \frac{1}{4}\times \frac{3}{4}}$
=9I64$={\displaystyle \frac{9I}{64}}$
Imax=I4+9I64+2I4×9I64−−−−−−−√${\displaystyle I_{max}=\frac{I}{4}+\frac{9I}{64}+2\sqrt{\frac{I}{4}\times \frac{9I}{64}}}$
=I4+9I64+2(3I16)=I64(16+9+24)=49I64${\displaystyle =\frac{I}{4}+\frac{9I}{64}+2(\frac{3I}{16})=\frac{I}{64}(16+9+24)=\frac{49I}{64}}$
Imin=I4+9I64−2(3I16)=I64(16+9−24)=I64${\displaystyle Imin=\frac{I}{4}+\frac{9I}{64}-2(\frac{3I}{16})=\frac{I}{64}(16+9-24)=\frac{I}{64}}$
So, ImaxImin=49I/64I/64=491.${\displaystyle \frac{I_{max}}{I_{min}}=\frac{49I/64}{I/64}=\frac{49}{1}.}$

Example

Calculate the slit width and other paremaeters of YDSE

Example: A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Solution:
Wavelength of light beam, λ$\lambda$
Distance of the screen from the slit, D=1m$D=1m$
For first minima, n=1$n=1$
Distance between the slits is d$d$
Distance of the first minimum from the centre of the screen can be obtained as, x=2.5mm=2.5×10−3$x=2.5mm=2.5\times {10}^{-3}$
Now, nλ=xd/D$n\lambda =xd/D$
⇒d=nλD/x=1×500×10−9×12.5×10−3=0.2 mm$\Rightarrow d=n\lambda D/x={\displaystyle \frac{1\times 500\times {10}^{-9}\times 1}{2.5\times {10}^{-3}}}=0.2mm$
Therefore, the width of the slits is 0.2 mm.

Definition

Diffraction of light

If a plane wave is passed through a small hole, spherical waves are obtained on the other side as if the hole itself is a source sending waves in all directions. Such bending of waves from an obstacle or an opening is called diffraction.

Example

Position of fringes due to diffraction

Example: A beam of light of wavelength 600 nm from a distant source falls on a
single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen 2m away. Find the distance between the first dark fringes on either side of the central bright fringe.

Solution:
we know
bsinθ=nλ$b\sin \theta =n\lambda$
b$b$ is the slit width, n$n$ is 1$1$ for 1st$1^{st}$ dark fringe.
So, sinθ=1×600×10−9m1×10−3m=6×10−4$\sin \theta ={\displaystyle \frac{1\times 600\times {10}^{-9}m}{1\times {10}^{-3}m}}=6\times {10}^{-4}$
Now, sinθ=distance of 1st dark fringe from centerdistance of screen$\sin \theta ={\displaystyle \frac{distanceof{1}^{st}darkfringefromcenter}{distanceofscreen}}$
So, distance of 1st$1^{st}$ dark fringe from centre =6×10−4×2$=6\times {10}^{-4}\times 2$ =1.2 mm$=1.2mm$
As both are equidistant, we have: Distance between dark fringes =1.2 mm+1.2 mm$=1.2mm+1.2mm$ =2.4 mm

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