Physics Ray Optics and Optical Instruments

$A=90^{?}$

In prism,  $r_1+c=A$

$r_1=90^{?}-c$                  …(i)

$sinc=\frac{1}{\mu }\Rightarrow cosc=\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1·sin30^{?}=\mu sin\left(r_1\right)\Rightarrow 1*\frac{1}{2}=\mu *sin\left(90^{?}-c\right)$

$\frac{1}{2}=\mu *\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

On squaring  $\frac{1}{4}={\mu }^2-1$

$\Rightarrow {\mu }^2=\frac{5}{4}\Rightarrow \mu =\frac{\sqrt[]{5}}{2}$

for point B,  $\frac{1}{u}=-0.10c{m}^{-1},\frac{1}{v}=0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10}=\left(1.5-1\right)\left(\frac{2}{R}\right)\Rightarrow \frac{1}{R}=\frac{1}{10}$

R = 10 cm

A camera lens focuses the light onto a light-sensitive surface like film or a digital sensor for capturing an image. The camera lens gathers as much light as possible for forming a clear and bright image. This lens focuses the collected light on a specific point like camera's film or its digital sensor. The focus is achieved due to the curvature of lens and the arrangement of its optical elements. This lens refracts light rays so that they converge at focal plane where film/sensor is located.

A telescope functions on the working principle of gathering light form distant source and then, forming image which can be magnified for observation. Based on their design, there are two types of telescopes including refracting telescopes and reflecting telescopes. A refracting telescope has a front lens called as objective lens. This front lens is a convex lens that collects light from distant objects and then bend it to the focal point. In a reflecting telescope, a concave mirror is the primary mirror that collects light and then, reflects it to a focal point.

At minimum deviation :

i = i'

r = r'

$r+r\text{'}=A\Rightarrow r=\frac{A}{2}$

Snell's law :

1 sin I = $\sqrt[]{3}sinr=\sqrt[]{3}sin\frac{i}{2}$ $\text{exception 25:}\text{exception 25:}\frac{}{}$

$\Rightarrow 2cos\frac{i}{2}=\sqrt[]{3}$

$\Rightarrow i=60°$

$\begin{array}{l}i=2r\Rightarrow r=30°\\ \Rightarrow A=60°\end{array}$

Since direction of incident ray (DIR) is from left to write, so considering refraction at point M, we can write

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$              

-> $\frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}$

$\Rightarrow -\frac{1.4}{v}=\frac{0.15}{25}+\frac{1.25}{40}=\frac{1.2+6.25}{200}=\frac{7.45}{200}$

$\Rightarrow v=-\frac{200*1.4}{7.45}=-37.58cm$

from given information : I = A = 60°, r = $\frac{A}{2}=30°$ =>  $\mu =\sqrt[]{3}=\frac{c}{v}$

$\Rightarrow v=\frac{c}{\sqrt[]{3}}\Rightarrow time=\frac{\left(\frac{5\sqrt[]{3}}{100}\right)}{\left(\frac{3*10^8}{\sqrt[]{3}}\right)}=5*10^{-10}s$

As observer is at O So height of water observed by observer

  $\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 - H) is height of observer

$So\frac{3H}{4}=17.5-H$          

$\frac{7H}{4}=17.5$            

7H = 70

H = 10

$m=\frac{f}{f+u}+\frac{-f}{-f-f}=+\frac{1}{2}$

$v=\frac{uf}{u+f}\Rightarrow v=\frac{-f*-f}{-f-f}=\frac{f^2}{-2f}=\frac{-f}{2}$

 

$\therefore$ dist of image from optical cnetre = $\frac{f}{2}$