Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A circular beam of light (diameter ) falls on a plane surface of a liquid. The angle of incidence

A circular beam of light (diameter \mathrm{=d}) falls on a plane surface of a liquid. The angle of incidence is \mathrm{45^{\circ}} and refractive index of the liquid is \mathrm{\mu }. The diameter of the refracted beam is:

Option: 1

\mathrm{d}


Option: 2

\mathrm{(\mu-1) \mathrm{d}}


Option: 3

\mathrm{\frac{\sqrt{2 \mu^2-1}}{\mathrm{~d}} \mathrm{~d}}


Option: 4

\mathrm{\frac{\sqrt{2 \mu^2-1}}{\mu} \mathrm{d}}


Answers (1)

best_answer

Form the figure, \mathrm{\mathrm{AB}=\frac{\mathrm{d}}{\cos \mathrm{i}}=\frac{\mathrm{d}^{\prime}}{\cos \mathrm{r}}}

\therefore \quad d^{\prime}=\frac{\cos r}{\cos i} d=\sqrt{2} \cos r \cdot d\ \ \ \ .............\mathrm{(i)} \\ Further,\ \quad \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mu}=\frac{1}{\sqrt{2} \mu}
Hence, \mathrm{\cos r=\sqrt{1-\frac{1}{2 \mu^2}}=\frac{\sqrt{2 \mu^2-1}}{\sqrt{2} \mu}}
Substituting in eq. \mathrm{(i)}, we get
\mathrm{\mathrm{d}^{\prime}=\frac{\sqrt{2 \mu^2-1}}{\mu} \cdot \mathrm{d}}

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

BSEB Super 50 performed ‘brilliantly’ in JEE Main, 23 scored above 95 percentile: Bihar Board chairman
anam-khan

IIIT Allahabad cut-offs for BTech ECE, IT, IT-Business Informatics goes up in last 3 years; seat matrix
anam-khan

IIIT Hyderabad JEE Main cut-offs for BTech CSE, ECE of last 5 years; modes of admissions for dual degrees

Latest Question