#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Playing with an old lens one morning, Ravi discovers that if he holds the lens 10cm away from a wall opposite to a window, he can see a sharp but upside -down picture of outside world on the wall. That evening, he covers alighted lamp with a piece of opaque paper on which he has pierced, a small hole 1 mm in diameter. by placing the lens between the illuminated card and the wall, he manages to produce a sharp image of diameter 5 mm on the wall.Answer the following questions based on the above information : (i)what is the power of the lens? (ii)in the evening experiment, how far away from the opaque paper did he place the lens?(iii)How far apart were the card and the wall?

Dronzer
15 Points
one year ago
According to question Ravi get a sharp image of object placed far away outside the window so image will be formed at focus and object is at infinity to get a sharp image.
1)  Now, v=f=10 cm   so P=100/10 P=10D

2) EVENING EXPERIMENT;

A/Q  m=hi/ho= -5= v/u .  So, v=-5u---(1)
1/f=1/v-1/u  =>  1/10=1/v-1/u-------(2)
From (1 )and (2) we get :
1/10=  -1/5u-1/u  =>  1/10=  -6/5u=>  5u= -60=>  u= -12 cm
So, (-12) cm far is the lens from the opaque paper.(front of
the lens)
3) v=? u= -12 cm and f=10 cm
1/f=1/v-1/u   =>  1/10= 1/v+1/12  => 1/10-1/12=1/v
=>6-5/60=1/v  =>  1/60=1/v   =>  v=60cm

So , card and the wall are 60 cm apart from each other.
(behind the lens)