Wave Optics Concept Page - 5

Example

Problems on points on screen with minimum and maximum intensity in YDSE

Example: A screen is placed 2 m$2m$ away from a single narrow slit. Find the slit width if the first minimum lies 5 mm$5mm$ on either side of the central maximum. (wave length =5000A∘$=5000A^{\circ }$)
Solution:
x=λ×Dω$x={\displaystyle \frac{\lambda \times D}{\omega }}$
⇒$\Rightarrow$ ω=λ×Dx$\omega ={\displaystyle \frac{\lambda \times D}{x}}$=5000×10−10×25×10−3$={\displaystyle \frac{5000\times {10}^{-10}\times 2}{5\times {10}^{-3}}}$=2×10−4 m$=2\times {10}^{-4}m$=0.02 cm$=0.02cm$

Example

Write intensity as a function of height from centre of screen in YDSE for two coherent light sources of equal intensity

Example: A point source is emitting light of wavelength 6000$6000$ Ao$\stackrel{o}{A}$ is placed at a very small height h$h$ above a flat reflecting surface MN as shown in the figure. The intensity of the reflected light is 36%$36\mathrm{\%}$ of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D$D$ from it. Find the shape of the interference fringes, on the screen.

Solution:
S′$S^{\prime }$ represents the another point source forms due to reflection from the mirror.
S′P=(x+h)2+y2−−−−−−−−−−−√$S^{\prime }P=\sqrt{(x+h{)}^2+y^2}$     and   SP=(x−h)2+y2−−−−−−−−−−−√$SP=\sqrt{(x-h{)}^2+y^2}$
Let path difference at point P  be Δ$\mathrm{\Delta }$
Using, Δ+SP=S′P$\mathrm{\Delta }+SP=S^{\prime }P$
Squaring and adding,  Δ2+4hx=−2Δ(x−h)2+y2−−−−−−−−−−−√${\mathrm{\Delta }}^2+4hx=-2\mathrm{\Delta }\sqrt{(x-h{)}^2+y^2}$
Squaring and adding again, Δ4+16h2x2+8hxΔ2=4Δ2(x2+h2−2hx+y2)${\mathrm{\Delta }}^4+16h^2{x}^2+8hx{\mathrm{\Delta }}^2=4{\mathrm{\Delta }}^2(x^2+h^2-2hx+y^2)$
⟹4(Δ2−4h2)x2−16Δ2hx+y2+4Δ2h2−Δ4=0$⟹4({\mathrm{\Delta }}^2-4h^2)x^2-16{\mathrm{\Delta }}^{2}hx+y^2+4{\mathrm{\Delta }}^2{h}^2-{\mathrm{\Delta }}^4=0$
which represents equation of circle as X2+Y2=R2$X^2+Y^2=R^2$
Thus the fringes will be of Circular shape centered at O.

Example

Problem on maximum and minimum intensity ratio calculation during interference

Example:
A point source is emitting light of wavelength 6000Ao$6000\stackrel{o}{A}$ is placed at a very small height h$h$ above a flat reflecting surface MN as shown in the figure. The intensity of the reflected light is 36%$36\mathrm{\%}$ of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D$D$ from it. Find the ratio of maximum to minimum intensities at P.

Solution:
Before reflection, intensity of light is Io$I_o$(say) and after reflection it becomes 0.36Io${0.36I}_o$
Imax=Io+0.36Io+2Io×0.36Io−−−−−−−−−√=2.56Io$I_{max}=I_o+{0.36I}_o+2\sqrt{I_o\times {0.36I}_o}=2.56I_o$
Imin=Io+0.36Io−2Io×0.36Io−−−−−−−−−√=0.16Io$I_{min}=I_o+{0.36I}_o-2\sqrt{I_o\times {0.36I}_o}=0.16I_o$
So, ImaxImin=25616=16:1${\displaystyle \frac{I_{max}}{I_{min}}}={\displaystyle \frac{256}{16}}=16:1$

Diagram

Intensity vs height on screen in YDSE

Diagram

Plot intensity vs height on screen in YDSE

Definition

Describe the importance of YDSE

Historically, Young's Double slit experiment, when it was first carried  out by Young in 1801, demonstrated that light was a wave, which seemed  to settle for once and all the debate whether light was corpuscular  (particle-like) or wave. Up until that time, Newton's corpuscular was  the prevailing view of light, in spite of alternate explanations such as  Huygens' wave front model (which had some serious shortcomings). After Young's demonstration of the wave properties of light, followed by work  by others such as Fresnel studying the interference and diffraction  properties of it, and especially after Maxwell's brilliant equations of  electromagnetism, physicists in the 19th century rejected Newton's  corpuscular theory and believed that light was a wave.

Example

Find the ratio of maximum to minimum itensity

Example: In Young's double slit experiment, first slit has width four times the width of the second slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe system.

Solution:

Let intensity from the smaller slit be I$I$. So, intensity from the bigger slit will be 4I$4I$ since it is 4$4$ times of smaller slit.
Now, Imax=4I+I+2(4I2)−−−−√$I_{max}=4I+I+2\sqrt{(4I^2)}$=5I+24I2−−−√$=5I+2\sqrt{4I^2}$=5I+2(2I)$=5I+2(2I)$=9I$=9I$
Imin=4I+I−2I(4I)−−−−√$Imin=4I+I-2\sqrt{I(4I)}$=5I−2(4I2)−−−−√$=5I-2\sqrt{(4I^2)}$=I.$=I.$
So, ImaxImin=9II=91.${\displaystyle \frac{I_{max}}{I_{min}}}={\displaystyle \frac{9I}{I}}={\displaystyle \frac{9}{1}}.$

Example

Shift of fringes when a slab/lens in introduced in front of a slit

Example: n$n$ transparent slabs of refractive index 1.5$1.5$ each, having thickness 1 cm,2 cm,..........$1cm,2cm,..........$ to n cm$ncm$ are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is 1 cm$1cm$ then find the value of n$n$.

Solution:
As the refractive index of all the slabs are 1.5$1.5$ so those may be a considered as a single slab with the thickness summed up.
Summed up thickness of n$n$ slabs Δt=1+2+3+...+n$\mathrm{\Delta }t=1+2+3+...+n$=n(n+1)2$={\displaystyle \frac{n(n+1)}{2}}$
Given, for small angle of incidence, shift of object: s=(1−1μ)Δt$s=(1-{\displaystyle \frac{1}{\mu }})\mathrm{\Delta }t$
so, ⇒1 cm=(1−11.5)n(n+1)2$\Rightarrow 1cm=(1-{\displaystyle \frac{1}{1.5}}){\displaystyle \frac{n(n+1)}{2}}$
⇒6=n(n+1)$\Rightarrow 6=n(n+1)$
⇒n2+n−6=0$\Rightarrow n^2+n-6=0$
⇒n2+3n−2n−6=0$\Rightarrow n^2+3n-2n-6=0$
⇒n(n+3)−2(n+3)=0$\Rightarrow n(n+3)-2(n+3)=0$
⇒(n+3)(n−2)=0$\Rightarrow (n+3)(n-2)=0$
So, n=−3$n=-3$ or n=2$n=2$
As, n$n$ can not be negative. so n=2$n=2$ 

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