Physics NCERT Exemplar Solutions Class 12th Chapter Nine

 $h=\frac{u^2}{2g}\Rightarrow u=\sqrt[]{2gh}$

$\frac{h}{3}=\sqrt[]{2gh}t-5t^2$

$5t^2-\sqrt[]{20h}t+\frac{h}{3}=0$

$t=\frac{\sqrt[]{20h}\pm \sqrt[]{20h-4*5*\frac{h}{3}}}{10}$

$\therefore \frac{t_1}{t_2}=\frac{\sqrt[]{20h}-\sqrt[]{\frac{40}{3}h}}{\sqrt[]{20h}+\sqrt[]{\frac{40}{3}h}}=\frac{1-\sqrt[]{\frac{2}{3}}}{1+\sqrt[]{\frac{2}{3}}}=\frac{\sqrt[]{3}-\sqrt[]{2}}{\sqrt[]{3}+\sqrt[]{2}}$

$T=k\sqrt[]{\frac{\rho r^3}{S^{3/2}}}$

$\left[T\right]$ ${\left[M{L}^{-3}\right]}^{1/2}{\left[L^3\right]}^{1/2}{\left[ML{T}^{-2}{L}^{-1}\right]}^{\frac{-3}{4}}$

$\left[T\right]$ $=\left[M^{1/2}{L}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{L}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{M}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{T}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=\left[M^{\frac{1}{4}}{T}^{3/2}\right]$

$R_1=R$  

$R_2=-R$       

$P=\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$               

$P=\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}-\left(\frac{-1}{R}\right)\right)$

$P=\frac{\left(\mu -1\right)2}{R}$              

L₂

R₁ = R

$R_2=\infty$               

$R_1=\infty$               

R₂ = ­­-R

Power of  $L_1=\frac{1}{f\text{'}\text{'}}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Power of $L_1=\frac{\left(\mu -1\right)*2}{R}=P$  

1° →  60'

⇒ 60' → 10°

$60\text{'}\to {\frac{\pi }{180}}^C$

$\theta =\frac{\pi *2*10^3}{180*3600}$          

L = 1.5 * 10¹¹ m

$D=L\theta =1.5*10^{11}*\frac{\pi }{180*3600}*2*10^3=1.45*10^{9}m$    

The image is real when it is formed by the actual convergence of rays and it can be projected on a screen. For example, when the object is placed beyond focus, and the image is formed either by the convex lens or concave mirror.
The image is virtual, if it is formed by the apparent divergence of rays and it cannot be captured on a screen. For example, when the image is formed by the convex mirror or plane mirror.

Simple microscope: When the object is placed within the focal length, the simple microscope works on the principle of producing an erect, virtual and magnified image. To magnify small objects, the simple microscope uses a single convex lens.
Compound microscope: It uses two lenses including an eyepiece and objective. The objective lens creates a real, magnified, and inverted image which acts as the object for the eyepiece, which in turn magnifies it further.

In Ray Optics And Optical Instruments questions, when using lens and mirror formulas, sign convention helps in maintaining consistency. The sign convention determines whether quantities like image distance, object distance, and focal length are taken as positive or negative. The incorrect usage of these sign conventions mostly leads to wrong magnifications and image positions. For accurate results, it is essential to follow the Cartesian sign.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) The length of the telescope tube L= f₀+f_e= 20+0.02= 20.02m

Also magnification is = 20/0.02= 1000

And image formed is inverted