Formation of image from a combination of lenses, mirrors and glass slabs
Example: A convex lens of focal length 40cm is held at a distance 12cm co-axially above a concave mirror of focal length 18cm. If the convex lens is replaced by a glass plate of thickness 6cm, refractive index μ=23 and gives rise to an image coincident with itself, then what will be the value of d?
Solution: When the ray from O passes through the slab of refractive index (μ), then there will be shift of point O to I1 and then this point I1 will act as source for the concave mirror. Shift = (1−μ1)t=(1−32)6 ⟹ shift =2cm, i.e., the object will appear to look closer by 2cm. Now as the final image is formed at a point O itself, so the ray from point I1 will retrace its own path (i-e, I1 should be at R of concave mirror). So, d−2+12=2×f1=2×18=36 ⟹d=26cm