Question

To find the focal length of a convex mirror, a student records the following data : The focal length of the convex lens is $f_1$ and that of mirror is $f_2$. Then taking index correction to be negligibly small, $f_1$ and $f_2$ are close to :

• A. $f_1=12.7 \, cm$ $f_2 = 7.8 \, cm$
• B. $f_1= 7.8 \, cm$ $f_2 = 12.7 \, cm$
• C. $f_1= 7.08 \, cm$ $f_2 = 25.4 \, cm$
• D. $f_1= 15.6 \, cm$ $f_2 = 25.4 \, cm$

Answer 1

Answer: (B)

$f_1= 7.8 \, cm$ $f_2 = 12.7 \, cm$

Answer 2

Take $f_{2}=12.07$ $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ $\frac{1}{12.7}=\frac{1}{25.4}+\frac{1}{u}$ $\frac{1}{12.7}=\frac{1}{25.4}=\frac{1}{u}$ $u=25.4=v '$ $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\frac{1}{f_{1}}=\frac{1}{25.4+13.6}+\frac{1}{10}$ $\frac{1}{f_{1}}=\frac{1}{39}+\frac{1}{10}$ $\frac{1}{f_{1}}=\frac{390}{49}=7.96$ The closest answers is (1) as option (3) and (4) are not possible.