A glass slab of thickness 2cm and refractive index 2 is placed in contact with a biconvex lens of focal length 20cm. The refractive index of the material of the lens 1.5 . The far side of the lens is silvered. Where should an object be placed in front of the slab so that it images on to itself?

Explanation

Step 1: Ray diagram: The path of a typical light ray is shown in Figure.
Step 2: The elements encountered are a glass slab, a silvered lens, and a glass slab.
Step 3: The lens is silvered, so it acts as a mirror. Since the image is formed on the object itself, the apparent position of the object after refraction from the glass slab should be at the center of curvature of the mirror.
Step 4: The effective focall length of the silvered lens is
`(1)/(f_(e))=(1)/(f_(m))-(2)/(f_(L))`
Now, `f_(L)=+20cm` ( assuming the +ve X-direction is to the right) .
It is a biconvex lens. Therefore,
`(1)/(f_(L))=(mu-1) ((1)/(R)+(1)/(R))` or `R=20cm`
Therefore, `f_(m)=-20//2=-10cm` and
` f_(e) =-5cm`
For the image to form on the object, the apparent position of the object after refraction at the glass slab must be at `2xxf_(e)=-10cm` , i.e., the object must appear to be 10cm to the left of the silvered lens.
Step 5: For the glass slab:
`Thickness =2cm , mu=2`
Shift due to slab `=s=t(1-(1)/(mu))=1cm`
Direction of shift is toward the left in the direction of the incident ray.
Therefore, the object should be at `10+1=11cm` to the left of the mirror so that after refraction at the glass slab it appears to be at 10 cm from the lens which is the center of curvature of the silvered lens. Hence, the rays retrace their paths and the image is on the object itself.