Question

A thin lens of a glass of a focal length 10 cm is immersed in water the new focal length is

Answer 1

Hello Dear.

Speed of the light in glass = 2 × 10⁸ m/s.
Speed of the light in air = 3 × 10⁸ m/s.

∵ Refractive Index of the glass with respect to the air = Speed of the light in air ÷ Speed of the light in Glass.
∴  μ of glass with respect to air = (3 × 10⁸) ÷ (2 × 10⁸)
     = 1.5

Now,

When the Lens is immersed in water, the Refractive Index of the Lens will also changes.

We know, The Refractive Index of the Water = 4/3.

Now,
Using the Lens Makers Formula,
 
$\frac{1}{f} = (n - 1)( \frac{1}{R1} - \frac{1}{R2} )$

Where, 
n = Refractive Index of the Medium with Respect to the air.
R₁ and R₂ are the Radius of the Curvatures of the Two Sides of the Lens.


Now,
In First Case,
Focal Length(f) = 10 cm.
Refractive Index(μ) = 1.5

∴  $\frac{1}{10} = (1.5 - 1)( \frac{1}{R1} - \frac{1}{R2} )$ 
 ⇒ $\frac{1}{10} = (0.5)( \frac{1}{R1} - \frac{1}{R2} )$
Taking it as eq(i).


Now,
In Second Case,
Focal Length = f
Refractive Index(μ) = 4/3

∴ $\frac{1}{f} = ( \frac{4}{3} - 1)( \frac{1}{R1} - \frac{1}{R2} )$
⇒ $\frac{1}{f} = ( \frac{1}{3} )( \frac{1}{R1} - \frac{1}{R2} )$

Taking it as eq(ii),

Now, Dividing the eq(i) by the eq(ii),
We get,

$\frac{f}{10} = \frac{0.5}{ \frac{1}{3} }$
⇒ f = 50 × 3
⇒ f = 150 cm.


∴ Focal Length of the lens when it is placed in water is 150 cm.


Hope it helps.

Answer 2

Answer:

Explanation: If the refractive index of glass ( from which lens is made) is 1.5 i.e., 3/2 and that of water is 4/3

Then new focal length F will be

F = 4f

For example: f = 10 cm then after dipping in water new focal length F is 40

Thankyou