A lens is made of three thin different mediums. Radius of curvature and refractive index of each medium is shown iin Figure., Surface AB is straight. An object is placed at some distance from the lens by which a real image is formed on the screen placed at a distance of 10cm from the lens. Find

a. The distance of the object from the lens.
b. A slab of thiskness `1.5cm` and refractive index 1.5 is introduced between the image and the lens. Find the new position of the object so that image is again formed on the screen.
c. Find the position of the image when same slabe is placed on the other side of the lens {as in part (b)}

Answer

(a) object is at infinity. (b) 190cm (c) there will not be any change in the image formation.

Explanation

a. `(1)/(f_(1))=(1.6-1)[(1)/(20)-(1)/(-30)]=(1)/(20)cm^(-1)`
`(1)/(f_(2))=(1.5-1)[(4)/(-30)-(1)/(15)]=(1)/(20)cm^(-1)`
`(1)/(f_(3))=(2.5-1)[(1)/(15)-(1)/(oo)](1)/(10)cm^(-1)`
`rArr (1)/(f_(eq)) =(1)/(20)-(1)/(20)+(1)/(10)=(1)/(10)`
`rArr f_(v)=10cm`
Since the image is formed on the screen which is 10 cm from the lens, so the object is at infinity.
b. Shift in the image away fromt he slab due to introduction of slab of
`Deltax=(1-(1)/(n))t=(1-(1)/(1.5))xx1.5=0.5`
The distance of the object so tha t image is again formed on the screen
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr (1)/(9.5)-(1)/(u)=(1)/(10)rArr u=190cm`
c. When the slab is placed on the other side of the lens, there will not be any change in the formation.