Question

For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

Answer 1

Given, 

u = 2.4 m, v = 12 cm, μ = 1.5, t = 1 cm

The shift produced by the glass plate is

\(d=t(1- \frac1\mu)= 1 \times (1-\frac1{1.5}) = \frac13 \ cm\)

So, final image must be produced at \((12 - \frac13)cm = \frac{35}3 \ cm\), from lens so that glass plate must shift it to produce image at screen.

So, we have

\(\frac1v - \frac1u = \frac1f\)

or, \(\frac{1}{12} - \frac1{-240} = \frac1f = \frac1{\frac{35}3} - \frac1u\)

or, \(\frac1u =\frac3{35}-\frac1{12}-\frac1{240}\)

\(= \frac{144-140-7}{1680}\)

\(=\frac{-3}{1680}\)

\(=\frac{-1}{560}\) cm

or, u = -560 cm = -5.6 m

Therefore, object should be shifted to = 5.6 - 2.4 = 3.2 m, for sharp image.