Question

A thin lens of glass $(\mu =1.5)$ of focal length $+10cm$ is immersed in water $(\mu =1.33)$. The new focal length is

• A. $20cm$
• B. $40cm$
• C. $48cm$
• D. $12cm$

Answer 1

Answer: (B)

$40cm$

Here, fa=+10cmfa=+10cm
μg=1.5=3/2,μw=1.33=4/3μg=1.5=3/2,μw=1.33=4/3
According to lens maker's formula
1f=(μ−1)(1R1−1R2)1f=(μ−1)(1R1−1R2) with μ=μLμMμ=μLμM
∴1fa=(μgμa−1)(1R1−1R2)∴1fa=(μgμa−1)(1R1−1R2)
=((3/2)1−1)C=((3/2)1−1)C
where C=(1R1−1R2)C=(1R1−1R2)
or 1fa=C2...(i)1fa=C2...(i)
also, 1fw=((3/2)(4/3)−1)C1fw=((3/2)(4/3)−1)C
1fw=C8...(ii)1fw=C8...(ii)
From (i) and (ii) we get
fafw=4fafw=4
or fw=4fa=4×10cm=40cm