Question

a concave spherical surface of refractive index 3/2 is immersed in water of refractive index 4/3 if a point object lies in water at a distance of 10 cm from the pole of the reflective surface calculate the position of the image give the radius of curvature of a spherical surface is 18cm
​

Answer 1

Answer:

Given :   R=−10 cm             n2=3/2               n1=4/3

Let the object is placed at a distance x in front of the spherical surface  i.e.  u=−x

Using:    vn2−un1=Rn2−n1

∴    v3/2−−x4/3=−103/2−4/3 

⟹  v=0.017x+1.33−1.5x

⟹   v<0  i.e. the image formed is always virtual in nature.

Answer 2

Answer:

v

h

2

−

u

h

1

=

R

h

2

−h

1

Putting h

2

=1.5,h

1

=1

u=−30cm.R=+20cm

So we get

v

1.5

−

−30

1

=

20

1.5−1

=

40

1

⇒

v

1.5

=

40

1

−

30

1

=

120

3−4

=

120

−1

⇒v=−180cm

i.e in the air side