Two identical plano-convex lenses `L_(1)(mu_(1)-1.4)` and `L_(2)(mu_(2)-1.5)` of radii of curvature `R=20cm` are placed as shown in Figure.

Q. Now, the secocnd lens is shifted vertically downward by a small distance 4.5mm and the extended parts of `L_(1)` and `L_(2)` are blackened as shown in figure. Find the new position of the image of the parallel beam.

Options

A. `200//9` cm behind the lens 2.5 mm below the principal axis of `L_(1)`
B. `100//9cm` in front of th elens 2mm below the prinicipal axis of `L_(1)`
C. `200//9cm` in front of the lens 2.5mm below the principal axis of `L_(1)`
D. `100//9`cm behind the lens 2mm below the princiipal axis of `L_(1)`

Answer

a.

Explanation

a. Focal lengths of lenses `L_(1)` and `L_(2)` are, respectively, given by
`(1)/(f_(1))=(mu-1)((1)/(R)-(1)/(oo))rArrf_(1)=50cm`
`(1)/(f_(2))=(mu-1)[(1)/(oo)-(-(1)/(R))]rArrf_(2)=40cm`
The equivalent focal length f of the combination is given by
`(1)/(f)=(1)/(f_(1))+(1)/(f_(2))rArrf=(200)/(9)cm`
Hence, the image of the parallel beam is formed on the common principal axis at a distance of 22.22cm from the combination on the right side.
b. Image formed by `L_(1)` is at a distance of 50cm behind the lens. This image lies on the principal axis of `L_(1)` and will act as an object for `L_(2)`
For `L_(2)`, object distance, `u=+50cm`
`f_(2)=+40cm`
`(1)/(upsilon)-(1)/(u)=(1)/(f)rArrmu=(200)/(9) cm`
Magnification caused by `L_(2), m=(upsilon)/(u)=(4)/(9)`
For `L_(2)` object `I_(1)` is at a distance of 4.5mm above its principal axis.
Hence, distance of image `I_(2)` of the object (virtual) `I_(1)` is at a distance `(4//9)xx4.5=2mm` above the principal axis of `L_(2)`
`[:. "height of image" =m xx "height of object" ]`
Hence, final image is at a distance of 22.22cm behind the combination at a distance of 2.5mm below the principal axis of `L_(1)`.