Question

Assertion: A double convex lens ($\mu$ = 1.5) has focal length 10 em, When the lens is immersed in water ($\mu$ = 4/3) its focal length becomes 40 cm.
Reason: $\frac{1}{f}=\frac{{\mu }_1-{\mu }_m}{{\mu }_m}=(\frac{1}{R_1}-\frac{1}{R_2})$.

• A. Both assertion and reason are true but the reason is the correct explanation of assertion
• B. Both assertion and reason are true but the reason is not the correct explanation of assertion
• C. Assertion is true but reason is false
• D. Both the assertion and reason are false
• E. Reason is true but assertion is false

Answer 1

Answer: (A)

Both assertion and reason are true but the reason is the correct explanation of assertion

Given: a double convex lens($\mu =1.5$) has focal length 10 cm, the lens is immersed in water $(\mu =\frac{4}{3})$

Solution:

If focal length of the lens when immersed in water be $f_w$

Focal length in air can be calculated by the formula,

$\frac{1}{f_a}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})⟹\frac{1}{f_a}=(\frac{1.5}{1}-1)(\frac{1}{R}-\frac{1}{-R})⟹\frac{1}{f_a}=\left(0.5\right)\times \frac{2}{R}⟹f_a=R..........\left(i\right)$

But given the focal length of lens in air , $f_a=10cm$

Hence , $R=10cm$

Focal length in water can be calculated by the formula,

$\frac{1}{f_w}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})⟹\frac{1}{f_n}=(\frac{1.5}{\frac{4}{3}}-1)(\frac{1}{R}-\frac{1}{-R})⟹\frac{1}{f_w}=(\frac{4.5}{4}-1)\left(\frac{2}{R}\right)⟹\frac{1}{f_w}=0.125\times \frac{2}{10}⟹f_w=40cm$

will be the focal length of the convex lens when immersed in water.