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Question

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is given by the expression
$$S_{1}^{\prime}=\left(\dfrac{n_{2}-n_{1}}{n_{2}+n_{1}}\right)^{2} S_{1}$$
In this equation, $$ S_{1} $$ represents the average magnitude of the Poynting vector in the incident light (the incident intensity), $$ S_{1}^{\prime} $$ is the reflected intensity, and $$ n_{1} $$ and $$ n_{2} $$ are the refractive indices of the two media.
(a) What fraction of the incident intensity is reflected for $$ 589-\mathrm{nm} $$ light normally incident on an interface between air and crown glass?
(b) Does it matter in part(a) whether the light is in the air or in the glass as it strikes the interface?


Solution

(a) The fraction reflected is
$$\dfrac{S_{1}^{\prime}}{S_{1}}=\left[\dfrac{n_{2}-n_{1}}{n_{2}+n_{1}}\right]^{2}=\left[\dfrac{1.52-1.00}{1.52+1.00}\right]^{2}=0.0426$$

(b) If medium 1 is glass and medium 2 is air,
$$\dfrac{S_{1}^{\prime}}{S_{1}}=\left[\dfrac{n_{2}-n_{1}}{n_{2}+n_{1}}\right]^{2}=\left[\dfrac{1.00-1.52}{1.00+1.52}\right]^{2}=0.0426$$
There is $$  { \text { no difference } } $$

Physics
NCERT
Standard XII

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