Get Answers to all your Questions

header-bg qa

When a ray of light falls on a prism of angle 48^{\circ}, it suffers minimum deviation. Refractive index of the material of prism is 1.6. The angle of incidence is nearly:

Option: 1

16^{\circ}


Option: 2

48^{\circ}


Option: 3

30^{\circ}


Option: 4

41^{\circ}


Answers (1)

best_answer

Given, \mathrm{A=48^{\circ},\mu =1.6}
For minimum deviation, \mathrm{r=\frac{A}{2}}
According to Snell’s law, \mathrm{\mu =\frac{sin\ i}{sin\ r}}
\mathrm{\begin{aligned} & \text { or } \quad \sin \mathrm{i}=\mu \sin \mathrm{r}=\mu \sin \frac{\mathrm{A}}{2}=1.6 \times \sin \frac{48^{\circ}}{2}=1.6 \times \sin 24^{\circ}=1.6 \times 0.407=0.651 \\ & \therefore \quad \mathrm{i}=\sin ^{-1}(0.651)=40.6^{\circ} \approx 41^{\circ} \end{aligned}}

Posted by

Nehul

View full answer

JEE Main high-scoring chapters and topics

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