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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Ray Optics

BITSAT 2016 · Physics · Ray Optics

For the angle of minimum deviation of a prism to be equal to its...

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

Options

  1. Alies between 2\sqrt{2} and 1
  2. Blies between 2 and 2\sqrt{2}
  3. Cis less than 1
  4. Dis greater than 2

Correct answer

B. lies between 2 and 2\sqrt{2}

Step-by-step solution

The angle of minimum deviation is given as \delta_{\min }=i+e-A for minimum deviation δmin⁡=A then 2A=i+e\begin{aligned} \delta_{\min }&=A \text{ then }\\ 2 A &=i+e \end{aligned} in case of\delta_{\min } i=e 2A=2ir1=r2=A2i=A=90∘\begin{aligned}2 A &=2 i \quad r_{1}=r_{2}=\frac{A}{2}\\ i &=A=90^{{}^{\circ}} \end{aligned}fromfromsmell's law 1sin⁡i=nsin⁡r1sin⁡A=nsin⁡A22sin⁡A2cos⁡A2=nsin⁡A22cos⁡A2=n\begin{array}{l}1\sin i=n \sin r_{1} \\ \sin A=n \sin \frac{A}{2}\\ 2 \sin \frac{A}{2} \cos \frac{A}{2}=n \sin \frac{A}{2}\\ 2 \cos \frac{A}{2}=n \end{array} when A=90∘=imin⁡ then nmin⁡=2i=A=0nmax⁡=2\begin{array}{l}\text{ when } \quad A=90^{{}^{\circ}}=i_{\min } \\ \text{ then } \quad n_{\min }=\sqrt{2}\\ i=A=0 \quad n_{\max }=2 \end{array}$
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