A glass sphere of index 1.5 and radius 40 cm has half its hemispherical surface silvered.
The point where a parallel beam of light, coming along a diameter, will focus (or appear to) after coming out of sphere, will be :

Options

A. 10 cm to the left of centre
B. 30 cm to the left of centre
C. 50 cm to the left of centre
D. 60 cm to the left of centre

Answer

D

Explanation

` R= 40 cm `
For refracting surface
`( mu _ 2 ) / ( v ) - ( mu _ 1 ) /( u ) = ( mu _ 2 - mu _ 1 ) /( R ) , n = 120 ` cm
for mirror ` (2 ) rArr ( 1 ) /( v ) + ( 1 ) /( u) = ( 1 ) /( f ) rArr ( 1 )/(v ) + ( 1 ) /( 40 ) = (1)/((-20))`
` rArr v = - 40//3 `
for refracting surface `" " `...(i)
` u = 80 - (40 ) /( 3 ) = (200) /( 3) rArr ( 1 ) /( v ) - ( 1.5 )/( - (200// 3 )) = ( 1 - 1.5 )/( - 40 ) `
` rArr ( 1 ) /( v ) + (4.5 ) /( 200) = ( 0.5 ) /( 40 ) `
` ( 1 ) /( v ) = ( 1 ) /(80) - ( 4.5 ) /(200) rArr v = - 100 ` cm
So the distance of image frosm the centre e will be 100 - 40 = 60 cm (left of the centre )