Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • (a) Derive lens maker’s formula for a biconvex lens. (b) A point object is placed at a distance of on the principal axis of a convex lens of

(a) Derive lens maker’s formula for a biconvex lens.

(b) A point object is placed at a distance of 12\; cm on the principal axis of a convex lens of focal length 10\; cm. A convex mirror is placed coaxially on the other side of the lens at a distance of 10\; cm. If the final image coincides with the object, sketch the ray diagram and find the focal length of the convex mirror.

 

 

 

 
 
 
 
 

Answers (1)

a)

consider a spherical surface with centre of curvature at c and radius of curvature R.

i is very small and the curved portion considered is the part of a large circle
from \Delta OMN
\tan < NOM= \frac{MN}{-u}---(1)
From \Delta MNC
\tan < NCM= \frac{MN}{R}---(2)
from \Delta MNI
\tan < NIM= \frac{MN}{v}---(3)
since the angle is very small
i= < NOM+< NCM= \frac{MN}{-u}+\frac{MN}{R}---(4)
< NCM= < NIM+r
r= < NCM-< NIM
= \frac{MN}{R}-\frac{MN}{v}---(5)
from snells law
n_{1}\sin i= n_{2}\sin r
n_{1}i= n_{2} r---(6)
from (5), (6) & (4)
n_{1}\left ( \frac{MN}{-u} +\frac{MN}{R}\right )= n_{2}\left ( \frac{MN}{R} -\frac{MN}{v}\right )
\frac{-n_{1}}{u}+\frac{n_{2}}{v}= \frac{n_{2}}{R}-\frac{n_{1}}{R}
or
\frac{n_{2}-n_{1}}{R}= \frac{n_{2}}{v}- \frac{n_{1}}{u}
\frac{n_{2}}{v}- \frac{n_{1}}{u}= \frac{n_{2}-n_{1}}{R}---(7)
A spherical lens can be considered as a two spherical surfaces The image of left surface act as virtual object for the other surface

for surface ABC

Apply equation (7) for surface ABC
\frac{n_{1}}{OB}+\frac{n_{2}}{BI_{1}}= \frac{n_{2}-n_{1}}{BC_{1}}---(8)
for surface ADC

I1 act as a virtual object for ADC
\frac{-n_{2}}{DI_{1}}+\frac{n_{1}}{DI}= \frac{n_{2}-n_{1}}{DC_{2}}---(9)\left ( from(7) \right )
for thin lens
BI_{1}= DI_{1},DC_{2}= R_{2}
adding (8) & (9)
\frac{n_{1}}{OB}+\frac{n_{1}}{DI}= (n_{2}-n_{1})\left ( \frac{1}{BC_{1}}+\frac{1}{DC_{2}} \right )
Suppose the object is at infinity
OB\rightarrow \infty ,DI= f
\Rightarrow \frac{n_{1}}{f}= (n_{2}-n_{1})\left ( \frac{1}{BC_{1}}+\frac{1}{DC_{2}} \right )
BC_{1}= R_{1},DC_{2}= -R_{2}
\Rightarrow \frac{n_{1}}{f}=\left ( n_{2}-n_{1} \right )\left [ \frac{1}{R_{1}} -\frac{1}{R_{2}}\right ]

b) 

For the lens

\\u=-12cm\\f=10cm\\for \ lens\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\\\frac{1}{v}=\frac{1}{10}-\frac{1}{12}\\\\v=60cm

As the final image is at the object itself the rays retrace its path. Therefore the image I' is at the centre of curvature of convex mirror and the radius of curvature= 60-10=50cm

Therefore the focal length of the convex mirror=50/2=25cm

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question