Ray Optics: Edge Cases and Subtle Traps (1)

easy • 3 min read
Tags Ray Optics

Question

A point object is placed 30 cm30 \text{ cm} in front of a concave mirror of focal length 20 cm20 \text{ cm}. A glass slab of thickness 6 cm6 \text{ cm} and refractive index 1.51.5 is then inserted between the object and the mirror. Find the new image position.

Solution — Step by Step

Using the mirror formula with sign convention (distances measured from pole, against incident light is negative):

1v+1u=1f,u=−30,f=−20\tfrac{1}{v} + \tfrac{1}{u} = \tfrac{1}{f}, \quad u = -30, \quad f = -20 1v=−120+130=−3+260=−160\tfrac{1}{v} = -\tfrac{1}{20} + \tfrac{1}{30} = \tfrac{-3 + 2}{60} = -\tfrac{1}{60}

So v=−60v = -60 cm — real image 6060 cm in front of mirror.

A slab of thickness tt and refractive index nn produces an apparent shift of the object toward the slab by:

shift=t(1−1n)=6(1−11.5)=6×13=2 cm\text{shift} = t\left(1 - \tfrac{1}{n}\right) = 6\left(1 - \tfrac{1}{1.5}\right) = 6 \times \tfrac{1}{3} = 2 \text{ cm}

The object appears closer to the mirror by 22 cm.

Apparent object distance: u′=−(30−2)=−28u' = -(30 - 2) = -28 cm.

 1v=1f−1u′=−120+128\tfrac{1}{v} = \tfrac{1}{f} - \tfrac{1}{u'} = -\tfrac{1}{20} + \tfrac{1}{28} 1v=−28+20560=−8560=−170\tfrac{1}{v} = \tfrac{-28 + 20}{560} = -\tfrac{8}{560} = -\tfrac{1}{70}

So v=−70v = -70 cm.

The reflected light also passes through the slab once more. The image formed by the mirror at 7070 cm is then shifted again toward the slab by another 22 cm — this time the actual image appears at 70−2=6870 - 2 = 68 cm from the mirror.

Final answer: Image is 68 cm\mathbf{68 \text{ cm}} in front of the mirror, real and inverted.

Why This Works

The slab doesn’t bend rays (parallel sides), but it lateral-shifts them. For paraxial rays, this is equivalent to seeing the object closer to the slab by t(1−1/n)t(1 - 1/n).

The subtlety: light passes the slab twice — once going to the mirror, once coming back. Both shifts must be applied for the final image position.

Alternative Method

Treat the slab using normal-shift on each leg. We could also compute it using u′=u+shiftu' = u + \text{shift} (object side) and v′=v−shiftv' = v - \text{shift} (image side) — same result, less confusion.

Common Mistake

Applying the slab shift only once. The slab is in the path of both incident and reflected rays, so the shift acts twice. Missing the second shift gives an answer of 7070 cm — a popular wrong option in JEE.

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