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`        A point object lies inside a transparent solid sphere of radius 20 cm and of refractive index n=2. When the object is viewed from the air through the nearest surface, it is seen at a distance 5 cm from the surface. Find the apparent distance of the object when it is seen through the farthest curved surface.`
7 years ago

28 Points
```										Dear Rohan,
Ans:- From the basic formulae of refraction we know that,
n2/v - n1/u = (n2 - n1)/R
Now it is given that v= -5cm. n2=1 n1=2
solving we get, u= - 40/3 cm
Now if we see from the other side, then u= - 80/3 cm. v=? and R=+20 cm
solving we get, v=-8 cm. Hence we will see it at a distance of 8 cm from that surface.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Rohan!!!

Regards,
SOUMYAJIT IIT_KHARAGPUR
```
7 years ago
Raunaq Mehta
39 Points
```										From the basic formulae of refraction we know that,n2/v - n1/u = (n2 - n1)/R Now it is given that v= -5cm. n2=1 n1=2solving we get, u= - 40/3 cmNow if we see from the other side, then u= - 80/3 cm. v=? and R=+20 cmsolving we get, v=-8 cm. Hence we will see it at a distance of 8 cm from that surface.
```
one year ago
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