A beam of light is converging towards a point $L$ . A plane parallel plate of glass of thickness $t$ , refractive index $μ$ is introduced in the path of the beam. The convergent point is shifted by :
(assume near normal incidence).

$t (1 - \frac{1}{μ}) a w a y$
$t (1 + \frac{1}{μ}) a w a y$
$t (1 - \frac{1}{μ}) n e a r e r$
$t (1 + \frac{1}{μ}) n e a r e r$

Question

A beam of light is converging towards a point $L$ . A plane parallel plate of glass of thickness $t$ , refractive index $μ$ is introduced in the path of the beam. The convergent point is shifted by :
(assume near normal incidence).

$t (1 - \frac{1}{μ}) a w a y$
$t (1 + \frac{1}{μ}) a w a y$
$t (1 - \frac{1}{μ}) n e a r e r$
$t (1 + \frac{1}{μ}) n e a r e r$

Toppr Admin · Accepted Answer

When glass plate of thickness t is introduced- the optical path increases by tn-xA0-So the convergence point shifts by t-x2013-t-x3BC- i-e- t-1-x2013-1-x3BC- nearer-xA0-

A beam of light is converging towards a point L. A plane parallel plate of glass of thickness t, refractive index μ is introduced in the path of the beam. The convergent point is shifted by : (assume near normal incidence).t(1−1μ)awayt(1+1μ)awayt(1−1μ)nearert(1+1μ)nearer

When glass plate of thickness t is introduced, the optical path increases by tn.So the convergence point shifts by t–tμ i.e. t[1–1μ] nearer.

When glass plate of thickness t is introduced, the optical path increases by $\frac{t}{n}$ .
So the convergence point shifts by $t - \frac{t}{μ}$ i.e. $t [1 - \frac{1}{μ}]$ nearer.