Question

4. A pond of depth 20 cm is half filled with an oil of
u= 1.4 and the other half is filled with water of
refractive index 1.33. Calculate apparent depth
of the tank when viewed normally.
[Ans. 14.66 cm]​

Answer 1

Given :

• Pond depth = 20 cm
• Pond is half filled with an oil of refractive index 1.4
• and the other half is filled with water of refractive index 1.33

To Find :

Apparent depth of the tank when viewed normally.

Theory:

When object in denser medium amd observer in rarer medium :

$\sf\:x_{apparent}=Virtual\:\:Depth=\dfrac{t}{\mu}$

Where ,

• t = Thickness of Denser medium
• and μ is refractive index of Denser medium

Solution :

We have to find apparent depth of the tank when viewed normally.

Given : Tank is half filled with water of refractive index 1.33 .

Let h be the portion of tank , filled with water. Then ,

$\sf\:u_1=1.33$

$\sf\:h=10cm$

And It's half filled with an oil of refractive index 1.4.

Let k be the portion of tank , filled with oil

Then ,

$\sf\:u_2=1.4$

$\sf\:h=10cm$

Now , Apparent depth of the tank :

$\sf\:Apparent\:Depth,x=\dfrac{h}{\mu_1}+\dfrac{k}{\mu_2}$

$\sf\implies\:x=\dfrac{10}{1.4}+\dfrac{10}{1.33}$

$\sf\implies\:x=\dfrac{10}{1.4}+\dfrac{10}{1.33}$

$\sf\implies\:x=\dfrac{100}{14}+\dfrac{1000}{133}$

$\sf\implies\:x=100[\dfrac{1}{14}+\dfrac{10}{133}]$

$\sf\implies\:x=100[\dfrac{133+140}{14\times133}]$

$\sf\implies\:x=100[\dfrac{273}{1862}]$

$\sf\implies\:x=100(0.1466)$

$\sf\implies\:x\approx\:14.66cm$

Therefore , Apparent depth of the tank when viewed normally is 14.66 cm [approx]