The convex surface of a thin concave-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place ? (b) If the concave part is filled with water (mu = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

Options

A. `Deltax=1.15cm, up`
B. `Deltax=3.15cm, down`
C. `Deltax=0.05cm, up`
D. `Deltax=0.15cm, down`

Answer

a.

Explanation

Method 1: When the concave part is filled with water of refractive index `4//3` , the optical arrangement is equivalent ot concave mirror of focal length F such that
`(1)/(F)=(1)/(f_(w))+(1)/(f_(g))+(1)/(f_(m))+(1)/(f_(g))+(1)/(f_(w))`
`(1)/(f_(w))=((4)/(3)-1)((1)/(60)-(1)/(oo))=(1)/(180)`
`f_(w)=180 cm`
`f_(g)=60cm` (calculated earlier)
`(1)/(F)=(1)/(180)+(1)/(60)+(1)/(10)+(1)/(60)+(1)/(180)=(26)/(180)`
`F+(180)/(26)cm`
`X_(1)=R=2F=(180)/(26)xx2=(180)/(13)=13.85cm`
`Deltax=15.0-13.85=1.15cm`
Method2: We use the equation
`(n_(2))/(x_(2))-(n_(1))/(x_(1))=(n_(2)-n_(1))/(R)`
For refraction at the interface '1' (air water),
`(4//3)/(x_(2))-(1)/(x_(1))=(4//3-1)/(oo)` (i)
The image of interface '1' is the object foro the interface '2'.
`(1.5)/(+20)-(1)/(x_(1))=(1.5-4//3)/(+60)`
`X_(1)=(360)/(26)=13.85`
`DeltaX=15.0-13.85=1.15cm`
Method 3: Using lensmaker's formula and the relation
`(1)/(F)=(1)/(x_(2))-(1)/(x_(1))`
`f_(m)=180cm` (using lensmaker's formula)
`f_(g)=60cm` (using lensmarker's formula)
`(1)/(-180)=(1)/(x_(2))-(1)/(x_(1))` (for the water lens) (i)
`(1)/(-60)=(1)/(+20)-(1)/(x_(1))` (for the glass lens) (ii)
(The image by the water by the glass lens is at `+20`, then the rays will fall normally on the mirrorgt)
Adding Eqs. (i) and (ii),
`(1)/(-180)+(1)/(-60)=(1)/(20)-(1)/(x_(1))`
`x_(1)=(180)/(13)=13.85cm`
`Deltax=15.0-13.85=1.15cm`