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## Homework Statement

A glass block of thickness 60 cm and refractive index 4/3 is placed in front of a concave mirror. A light source is then placed on the center of curvature of the concave mirror. Find the position of the final image formed. (Focal length is 30 cm)

## Homework Equations

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##Shift = d (1 - \frac{1}{n})##

## The Attempt at a Solution

I found two different results as final position of the image:

1. At the center of curvature. Because there is no enough space to shift for beams which coming from the source.

2. Its distance from concave mirror is 105 cm. (Calculations are below)

##Shift = 60-60.(\frac{3}{4})=15 cm##

At first the effective distance of object (light source) is shifted towards the mirror by an amount of 15 cm. Then, for concave mirror

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##\frac{1}{30} = \frac{1}{60-15} + \frac{1}{q} => q= 90 cm##

But since the beams will shift (15 cm), the final position of image will be at 90 +15 = 105 cm.

Which is correct, if any :(

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