Question

A thick lens has an index of refraction of 1.60. It is 4-cm thick and is surrounded by air. The radius of curvature of the front surface is 5 cm, and the radius of curvature of the back surface is 12 cm. A 1-cm tall object is placed to the left of the first surface. At what angle above the optic axis should a ray strike the front side of the lens 0.5 cm above the optical axis so that it emerges from the back side of the lens traveling parallel to the optical axis? Use matrix techniques.

Answer 1

Lets say we are taking diverging lens ->as image formed at infinity => R₁->-ve, R₂->+ve

Parallel to axis finally=> Image at infinity

₂/v-₁/u =(₂-₁)/R₂

length R is taken in same direction of direction of light =>R is +ve =+12 cm

Light going from lens to air (Remember we are approaching problem from the back i.e., backtracing because we know image distance =infinity, parallel to axis =>infinity image distance. and don't know object distance)

=>1/infinity -1.6/u = (1-1.6)/+12

=>u=-1.6*12/.6 =+32 cm => this is image distance after 1st reflection from the second surface

Thickness = 4 cm => image distance after 1st reflection from the first surface =32+4 =36 cm

And it is +ve because it is behind the lens, And the image distance direction is in the same direction of incoming light direction.=>+36 cm

1.6/36 -1/u= (1.6-1)/-5 => 1/u =0.6/5 +1.6/36 => u=6.09 cm => +ve => our assumption is wrong

so take convex surface at front and diverging surface at back

=>1.6/36-1/u =(1.6-1)/5 => 1/u =-0.6/5 +1.6/36 => u=-13.33 cm

Now apply trigonometry =>Vertical side =1-0.5 =0.5 cm

Horizontal side =u =13.33 cm

=>tan =0.5/13.33

and then apply trigonometry

(there might be some silly mistakes check it).