The distance travelled by a beam of light before it starts to spread out due to diffraction is called Fresnel distance.
Fresnel distance $Z_{f}=\frac{a^{2}}{\lambda}$
Where a = size of aperture, l= wavelength.
Ray optics is valid only when the wave length of light (l) tends to zero.
(Or) when Fresnel distance $Z_{f}$ tends to infinity.
Ans. The phenomenon of restricting the electric vectors flight into a single direction is called polarization.
Ans. The intensity ($I$) of polarized light transmitted through an analyzer is equal to the product of the incident intensity ($I_{circ}$)and the square of the cosine of the angle between the transmission axes of the polarizer and the analyzer.
$I=I_{\circ}cos^{2}\theta$
Ans. When light is incident at polarizing angle \(i_{p}\) at the interface of a refracting medium, the refractive index of the medium is equal to the tangent of the polarizing angle.
\(m_u=tan\,i_{p}\)
Ans. When a monochromatic light beam passing through a polarizer is incident on the surface of a prism at Brewstar angle, then total transmission of light takes place through the prism for a particular alignment of the polarizer.
Ans. Doppler Effect in Light: According to Doppler Effect, whenever there is a relative motion between a source of light andobserver, the apparent frequency of light received by the observer is different from the true frequency of light emittedactually from the source of light.
The apparent frequency of light increases when the distance between source of light and observer is decreasing and the apparent frequency of light decreases if the distance between source of light and observer is increasing.
Blue shift: When source and observer approach each other, △ υ is +ve, i.e., apparent frequency increases or apparent wavelength decreases. This is called "Blue shift".
Red shift: When source and observer recede away from each other, △ υ is +ve, i.e., apparent frequency increases or apparent
decreases or apparent wavelength increases. This is called "Red shift‟.
Ans. $$\begin{aligned}&\mathrm{Y_{1}=a\cos\omega t,~Y_{2}=a\cos(\omega t+\emptyset)}\\&\mathrm{Y=Y_{1}+Y_{2}=a\cos\omega t+a\cos(\omega t+\emptyset)}\\&\mathrm{Y=2a\cos}(\frac{\phi}{2})\mathrm{\cos}(\omega t+\frac{\phi}{2})\\\end{aligned}$$Amplitude of the resultant $A=2a\cos (\frac{\Phi}{2})$$$Intensity=I=A^{2}=4a^{2}\cos^{2}{(\frac{\phi}{2})}$$Condition for maximum intensity:$\phi = 0, \pm 2\pi , \pm 4\pi , \pm 6\pi , . . . . . . . . . . . . n( 2\pi )$ n=0,1,2,3,.......
$I_{\mathrm{max}}=4a^2$
Condition for minimum intensity
$\phi = 0, \pm \pi , \pm 3\pi , \pm 5\pi , . . . . . . . . . . . ( 2$n+ 1$) \pi$ n=0,1,2,3,....... $I_{\mathrm{min}}=0$
Ans. In figure all the black stripes should be of equal width, while the width of the intermediate white stripes should
increase as you go from the left to the right. For example let all black stripes have a width of $5 mm$. Let the width of the first two white stripes be $0.5 mm$ each, the next two white stripes be $1 mm$ each, the next two $1.5 mm $each etc. Paste this patternon a wall in a room or laboratory, at the height of your eye.Now watch the pattern, preferably with one eye. By moving away or closer to the wall, find the position where you can justsee some two black stripes as separate stripes. All the black stripes to the left of this stripe would merge into one another andwould not be distinguishable. On the other hand, the black stripes to the right of this would be more and more clearly visible.Note the width d of the white stripe which separates the two regions, and measure the distance $D$ of the wall from your eye.Then $d/D$ is the resolution of your eye.
Ans. Let I$_0$ be the intensity of polarized light after passing through first polarizer P$_1.$ Then the intensity of light after passing
through second polariser $P_2$ will be $I=I_0\cos^2\theta.$Where $\theta$ is the angle between $P_1$ and $P_2$ axes. Since $P_1$ and $P_3$ are crossed the angle between the $P_2$ and$ P_3$ will be $(90- \theta).$
Hence the intensity of light emerging from $P_3$ is$$I= I_{o}cos^{2}\theta. \cos^{2}\left ( 90- \theta \right )$$ $$I= I_{o}cos^{2}\theta. sin^{2} \theta$$$$I= \frac {I_{0}}4\sin^{2}2 \theta$$Therefore the transmitted intensity will be maximum when $\theta=45^0$