Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A small point object is inside a glass surface as shown. Refraction takes place at concave spherical boundry separating glass air medium. For the image to be real the object distance u is (R-Radius of curvature of the surface , $\mu of glass =\large\frac{3}{2}$ )


$(a)\;u=2R \\ (b)\;u < 3R \\ (c)\;u > 3R \\ (d)\;u < R$

Can you answer this question?

1 Answer

0 votes
We know that
$ \large\frac{ \mu_2}{v}=\frac{\mu _1}{u}=\frac{\mu_1-\mu_1}{R}$
Since object is in glass,
$\mu _2=1, \mu_1=1.5$
We have,
$\large\frac{1}{v} -\frac{1.5}{(-u)}=\frac{1-1.5}{-R}$
$\large\frac{1}{v}+\frac{3}{2u} =\frac{1}{2R}$
=> $\large\frac{1}{v} =\frac{1}{2R}-\frac{3}{2u}$
for image to be real,
v is to be positive
$\large\frac{1}{2R} > \frac{3}{2u}$
or $ u > 3R$
Hence c is the correct answer.
answered Jan 16, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App