Browse by Stream
QnA
Home

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • (a) Explain with reason, how the power of a diverging lens changes when (i) it is kept in a medium of refractive index greater than that of the lens. (ii) incident red light is replaced by violet light. (b) Three lenses L1, L2</s

(a) Explain with reason, how the power of a diverging lens changes when (i) it is kept in a medium of refractive index greater than that of the lens. (ii) incident red light is replaced by violet light.
(b) Three lenses L1, L2, L3 each of focal length 30 cm are placed co-axially as
shown in the figure. An object is held at 60 cm from the optic centre of lens L1.
The final real image is formed at the focus of L3. Calculate the separation
between (i) (L1and L2) and (ii) (L2 and L3).

 

 

 
 
 

Answers (1)

a)
(i) Power of a lens
p= \left ( \frac{n_{2}}{n_{1}}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )
for a diverging lens power is negative when the lens is immersed in a medium with refractive index > refractive index of a lens, then
\frac{n_{2}}{n_{1}}< 1 & power will become positive
So a diverging lens behaves as a converging lens when immersed in a medium of refractive index greater than that of the lens.
ii) when the incident light is replaced by a violet light wavelength decreases, therefore the refractive index increases, so power also increases.
b)
For L1
u= -60\, cm
f= 30\, cm
\frac{1}{f}= \frac{1}{v}-\frac{1}
\frac{1}{v}= \frac{1}{f}+\frac{1}{u}= \frac{1}{30}-\frac{1}{60}= \frac{30}{1800}
v= 60\, cm
Since the final image is formed at the focus of L3, the rays after L2 should be parallel to the principal axis, which happens when the rays are coming from the focus of L2

So the distance between
L1 & L2  is 90 cm
L2  & L3 can have any reasonable value

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question