Question

A thin convex lens made of glass n=1.5has a focal lenght of 20cm it is now completely immrsed inna liquid of n=1.75find new focal lenght of lens

Answer 1

Explanation:

Given A thin convex lens made of glass n=1.5 has a focal length of 20 cm it is now completely immersed in a liquid of n=1.75 find new focal length of lens

• The focal length of the glass in air is given by
• 1/fo = (n – 1) (1/R1 – 1/R2)
• 1/20 = (1.5 – 1) (1/R1 – 1/R2) ------------1
• The focal length of glass lens in liquid is given by
• 1/f l = (n – 1) (1/R1 – 1/R2)
• Now n = n1 / n2
•           = 1.5 / 1.75
•         = 6/7
• So 1/f l = (6/7 – 1)(1/R1 – 1/R2) ------------------2
• Dividing eqn 1 by 2 we get
• So fl / 20 = 0.5 / - 1/7
•               = - 3.5
• So f l = - 3.5 x 20
•          = - 70 cm
• Since the focal length of lens is negative, the nature of immersed lens is concave.

Answer 2

The new focal length of lens is 60 cm.

Given:

Focal length = 20 cm

Refractive index of lens = 1.5

Refractive index of liquid = 1.75

Solution:

According to lens maker's formula,

$\frac{1}{f_{1}}=\left[\frac{\left(n_{2}-n_{1}\right)}{n_{1}}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

Now, f₁ = 20 cm, n₁ = 1 (air) and n₂ = 1.5

$\frac{1}{20}=\left[\frac{(1.5-1)}{1}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{20}=(0.5)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \longrightarrow (1)$

$\frac{1}{f_{2}}=\left[\frac{\left(n_{2}-n_{1}\right)}{n_{1}}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

Now, n₁ = 1.75 and n₂ = 1.5

$\frac{1}{f_{2}}=\left[\frac{(1.5-1.75)}{1.75}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{f_{2}}=(-0.166)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \longrightarrow (2)$

$\frac{f_{2}}{20}=\frac{0.5}{(-0.166)}$

$f_{2}=\frac{10}{(-0.166)}$

$\therefore f_{2}= 60 \ cm$