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Effective Focal Length of Lens
Formation of image from a combination of lenses, mirrors and glass slabs
Example: A convex lens of focal length 40 cm is held at a distance 12 cm co-axially above a concave mirror of focal length 18 cm. If the convex lens is replaced by a glass plate of thickness 6 cm, refractive index μ=23​ and gives rise to an image coincident with itself, then what will be the value of d?
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 =2 cm, i.e., the object will appear to look closer by 2 cm.
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=26 cm
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Effective Focal Length
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