Class 12th

$\begin{array}{l}LetI={\displaystyle {\int }_{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^2}xdx.\\ I=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^2}xd(x)-----(i)\\ =2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^2}\left(\frac{\pi }{2}-x\right)dx\\ =2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}co{s}^2}xdx-----(ii)\end{array}$

$\begin{array}{l}\left(i\right)+\left(ii\right)weget,\\ ?{\displaystyle {\int }_{-a}^{a}f}(x)dx=2{\displaystyle {\int }_0^{a}f}(x)dxiff(-x)=f(x)\end{array}$

$\begin{array}{l}f\left(x\right) = si{n}^{2}x\hfill \end{array}$

$\begin{array}{ll} f\left(x\right) = si{n}^2\left(–x\right) = {\left(–1\right)}^{2}si{n}^{2}x= si{n}^{2}x\hfill & \hfill \end{array}$

$\begin{array}{l}if,f\left(x\right) =f\left(–x\right)\hfill \end{array}$

$\begin{array}{l}\left\{?{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\right\}\\ 2I=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(si{n}^{2}x+co{s}^{2}x\right)}dx\\ I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}1}dx=\frac{\pi }{2}.\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(}2logsinx-logsin2x)dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(logsi{n}^{2}x-logsin2x\right)}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{si{n}^{2}x}{sin2x}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{si{n}^{2}x}{2sinx·cosx}dx.\end{array}$

$\begin{array}{l}I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{1}{2}tanx\right)dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\frac{1}{2}tan\left(\frac{\pi }{2}-x\right)\right]}dx{\displaystyle {\int }_0^{a}f(x)dx=}{\displaystyle {\int }_0^{a}f(a-x)dx}\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(log\frac{1}{2}cotx\right)}dx-----(ii)\\ \left(i\right)+\left(ii\right)\\ I+I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\left(\frac{1}{2}tanx\right)+log\left(\frac{1}{2}cotx\right)\right]}dx\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{1}{2}tanx*\frac{1}{2}cotx\right)dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{1}{4}dx.\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(}log1-log4)dx\\ =0–log4{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x\\ =–log4*\frac{\pi }{2}\\ I=-\frac{log{2}^2*\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =-\frac{2log2*\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =\frac{-\pi }{2}log2.\end{array}$

Kindly go through the solution

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}(1+tanx)dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}\left[1+tan\left(\frac{\pi }{4}-x\right)\right]dx?\{{\displaystyle {\int }_0^{a}f(x)dx=}{\displaystyle {\int }_0^{a}f(a-x)dx}\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left[1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}\right]dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}·\left\{1+\frac{1-tanx}{1+tanx}\right\}dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left\{\frac{1+tanx+1-tanx}{1+tanx}\right\}dx\\ I={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left(\frac{2}{1+tanx}\right)dx-----(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ I+I={\displaystyle {\int }_0^{\frac{\pi }{4}}\left\{log(1+tanx)+log\left(\frac{2}{1+tanx}\right)\right\}}dx\\ I={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left\{\underset{\_}{1+tanx}*\frac{2}{1+tanx}\right\}dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}2dx\\ ={[xlog2]}_0^{\frac{\pi }{4}}\\ =log2\left[\frac{\pi }{4}-0\right]=\frac{\pi }{4}log2\\ I=\frac{\pi }{8}log2.\end{array}$

A single slit diffraction pattern produces a central maximum and diminishing side bands as it interacts with itself and results from the wavefront bending around the edges of the slit. However, the double slit interference pattern forms equally spaced bright and dark fringes, and it is due to the light superposition from two different coherent sources.

The following conditions are needed to observe sustained (stable) interference:

• The two sources should have a constant phase difference, i.e; they should be coherent.
• The light waves need to have almost the same frequencies.
• The sources must emit waves with comparable amplitudes.
• The path difference should be within the coherence length.

Young's Double-Slit Experiment demonstrates the interference phenomenon. It provides strong proof for the wave nature of light. This experiment shows how due to constructive and destructive interference, two coherent light sources create a pattern of bright and dark fringes.

$\begin{array}{l}Let\text{?}\text{?}\text{?}\text{?}I={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}(1+tanx)dx?????(i)\\ ={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}\left[1+tan\left(\frac{?}{4}?x\right)\right]dx?\{{\displaystyle {?}_0^{a}f(x)dx=}{\displaystyle {?}_0^{a}f(a?x)dx}\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left[1+\frac{tan\frac{?}{4}?tanx}{1+tan\frac{?}{4}tanx}\right]dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}·\left\{1+\frac{1?tanx}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\frac{1+tanx+1?tanx}{1+tanx}\right\}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left(\frac{2}{1+tanx}\right)dx?????(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ I+I={\displaystyle {?}_0^{\frac{?}{4}}\left\{log(1+tanx)+log\left(\frac{2}{1+tanx}\right)\right\}}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\underset{\_}{1+tanx}*\frac{2}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}2dx\\ ={[xlog2]}_0^{\frac{?}{4}}\\ =log2\left[\frac{?}{4}?0\right]=\frac{?}{4}log2\\ I=\frac{?}{8}log2.\end{array}$