logo

Ask Questions, Get Answers

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

A double convex lens $(\mu =1.5) $ has its focal length 10 cm. What will be its focal length if it is immersed in liquid of $(\mu_1=\large\frac{4}{3})$

$(a)\;\frac{10}{3}\;cm \\ (b)\;\frac{40}{3} \;cm \\ (c)\;40\;cm \\ (d)\;10\;cm $

Can you answer this question?
 
 

1 Answer

0 votes
focal length in air $\large\frac{1}{f_a}=(\mu_1 -1) \bigg[\large\frac{1}{r}-\frac{1}{-r} \bigg]$
$\qquad= \bigg(\large\frac{3}{2}-1 \bigg) \frac{2}{r}$
focal length in liquid $\large\frac{1}{fw} =\bigg(\large\frac{\mu_1}{\mu_2} -1\bigg) \bigg[ \large\frac{2}{r}\bigg]$
$\qquad= \bigg (\large\frac{3/2}{4/3} -1\bigg) \frac{2}{r}$
$\qquad= \bigg( \large\frac{9}{8} -1\bigg) \frac{2}{r}$
$\large\frac{f_w}{f_a} =\large\frac{(\Large\frac{3}{2}-1)}{(\Large\frac{9}{8}-1) }$
$\qquad= 4$
$\therefore f_w= f_a \times 4$
$\qquad= 10 \times 4$
$\qquad =40 \;cm$
Hence c is the correct answer.
answered Jan 23, 2014 by meena.p
 

Related questions

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