English
Karnataka Board PUCPUC Science Class 11

A Small Object is Embedded in a Glass Sphere (μ = 1.5) of Radius 5.0 Cm at a Distance 1.5 Cm Left to the Centre. Locate the Image of the Object as Seen by an Observer - Physics

Advertisements
Advertisements

Question

A small object is embedded in a glass sphere (μ = 1.5) of radius 5.0 cm at a distance 1.5 cm left to the centre. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere.

Sum
Advertisements

Solution

Given,
Radius of the sphere = 5.0 cm
Refractive index of the sphere (μ1) = 1.5
An object is embedded in the glass sphere 1.5 cm left to the centre.

(a)  
When the image is seen by observer from left of the sphere,
from surface the object distance (u) = − 3.5 cm
μ1 = 1.5
μ2 = 1
v1 = ?
Using lens equation: 
\[\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]
\[\frac{1}{v_1} - \frac{1 . 5}{- (3 . 5)} = \frac{- 0 . 5}{- 5}\]
\[\frac{1}{v_1} = \frac{1}{10} - \frac{3}{7}\]
\[= \frac{7 - 30}{70}\]
\[= \frac{- 23}{70}\]
\[v_1  = \frac{- 70}{23} \simeq  - 3  \text{ cm }\] 
So the image will be formed at 2 cm (5 cm - 3cm) left to centre. 

(b) When the image is seen by observer from the right of the sphere,
u = −(5.0 + 1.5) = − 6.5,
R = −5.00 cm
μ1 = 1.5, μ2 = 1, v = ?

Using lens equation:
\[\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\] 
\[\frac{1}{v} - \frac{1 . 5}{6 . 3} = \frac{1}{10}\] 
\[\frac{1}{v} = \frac{1}{10} - \frac{3}{13}\]
\[= \frac{13 - 30}{130}\]
\[= \frac{- 17}{130}\]
\[v = \frac{- 130}{17} =  - 7 . 65  \text{ cm }\]
Therefore, the image will be formed 7.6 − 5 = 2.6 towards left from centre.

shaalaa.com
  Is there an error in this question or solution?
Chapter 18: Geometrical Optics - Exercise [Page 415]

APPEARS IN

HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 18 Geometrical Optics
Exercise | Q 40 | Page 415
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×