Get Answers to all your Questions

header-bg qa

IMG_20190216_073603.jpg

A transparent solid cylindrical rod has a refractive index of \frac{2}{\sqrt{3}} It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.

The incident angle \Theta for which the light ray grazes along the wall of the rod is

Answers (1)

 Applying Snell's law at Q

                                     Q

\mu=\frac{sin90^{o}}{sin(90-\alpha)}=\frac{1}{cos\alpha}

cos\alpha=\frac{1}{\mu}

sin\alpha=\sqrt{1-cos^{2}\alpha}=\sqrt{1-\frac{1}{\mu^{2}}}=\sqrt{\frac{\mu^{2}-1}{\mu^{2}}}                                (i)

Snell's law at P

sin\theta=\mu sin\alpha=\sqrt{\mu^{2}-1}

sin\theta=\sqrt{\left ( \frac{2}{\sqrt{3}} \right )^{2}-1}=\sqrt{\frac{4}{3}-1}=\frac{1}{\sqrt{3}}

\theta=sin^{-1}\left(\frac{1}{\sqrt{3}} \right )

Posted by

Safeer PP

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE