Question

9. A convex lens of focal length 10cm is held horizontally 5cm above water surface. A point
object is placed 25cm above the lens. Find the position of image formed by refraction of light
by lens and water surface. Hw 4/3.
[15.52cm]
​

Answer 1

Given:

Focal length of convex lens = 10 cm , its kept 5cm above water surface.

Object distance from lens = 25 cm

Refractive Index of water is 4/3.

To find:

Position of image after refraction from lens and water.

Calculation:

Image produced from lens will acts as an object placed inside water at a particular depth. The final observer shall see the raised image of the image produced inside water.

Applying Lens Formula :

$\therefore \: \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$= > \: \dfrac{1}{10} = \dfrac{1}{v} - \dfrac{1}{( - 25)}$

$= > \: \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{25}$

$= > \: \dfrac{1}{v} = \dfrac{5 - 2}{50}$

$= > \: \dfrac{1}{v} = \dfrac{3}{50}$

$= > v = \dfrac{50}{3} = 16.67 \: cm$

Since the lens is placed 5 cm above the water surface , so the actual depth of the image inside water be d (let):

$\therefore \: d = 16.67 - 5 = 11.67 \: cm$

Now , this image will acts as the object and produce a raised image.

$\therefore \: apparent \: depth = \dfrac{real \: depth}{ \mu}$

$= > \: apparent \: depth = \dfrac{11.67}{ (\frac{4}{3}) }$

$= > \: apparent \: depth = 8.77 \: cm$

So the final image will be produced 8.77 cm from water surface.