Question

[img]https://upload.cc/i1/2020/04/20/ub5KAU.jpg[/img]



question in picture .R=4

Answer 1

As per the given question,

Let the focal length of the lens 1 is f1 and focal length of the lens 2 is f2,

$R_1=8cm, R_2=2cm, R_1'=\infty , R_2'=(10+R)$

R=4

a) We know that

$\dfrac{1}{f_1}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\dfrac{1}{f_1}=\dfrac{1}{-8}-\dfrac{1}{2}=\dfrac{-5}{8}$

So, $f_1=\dfrac{-8}{5}=-1.6cm$

Similarly,

$\dfrac{1}{f_2}=\dfrac{1}{R_1'}+\dfrac{1}{R_2'}=\dfrac{1}{\infty}+\dfrac{1}{-14}$

Hence focal length of the lens 2 will be $f_2=-14cm$

b) Net focal length $f_{net}$

$\dfrac{1}{f_{net}}=\dfrac{1}{f_1}+\dfrac{1}{f_2}+\dfrac{d}{f_1f_2}$

Now, substituting the values,

$\dfrac{1}{f_{net}}=\dfrac{1}{-1.6}+\dfrac{1}{-14}+\dfrac{18-12}{1.6\times 14}$

$\dfrac{1}{f_{net}}=-(\dfrac{14+1.6-6}{1.6\times 14})=\dfrac{-9.6}{22.4}=\dfrac{-3}{7}$

$f_{net}=\dfrac{-7}{3}cm$

Object is at origin, so u =12cm, let the final image is at v,

$v=\dfrac{f_{net}u}{f_{net}+u}=\dfrac{\dfrac{-7}{3}\times(-12)}{-12-\dfrac{-7}{3}}=\dfrac{84}{-43}=-1.95cm$

c) $M=v/u=-1.95/(-12)=0.1625$