In an experiment for the determination of the refractive index of glass of a prism by an $i - \delta$ plot,it was found that a ray incident at an angle of $35^o$ suffers a deviation of $40^o$ and emerges at an angle of $79^o$. In that case,which of the following is closest to the maximum possible value of the refractive index?

  • A
    $1.7$
  • B
    $1.8$
  • C
    $1.5$
  • D
    $1.6$

Explore More

Similar Questions

Which colour has more speed in a glass prism? Blue or yellow?

$A$ thin prism of angle $6^{\circ}$ and refractive index for yellow light $(n_{Y}) = 1.5$ is combined with another prism of angle $5^{\circ}$ and $n_{Y} = 1.55$. The combination produces no dispersion. The net average deviation $(\delta)$ produced by the combination is $(\frac{1}{x})^{\circ}$. The value of $x$ is.......

Three prisms $1, 2,$ and $3$ have a prism angle $A = 60^{\circ}$. Their refractive indices are $1.4, 1.5,$ and $1.6$ respectively. If $\delta_1, \delta_2,$ and $\delta_3$ are their angles of deviation,then:

$A$ ray of light is incident on the first face of an equilateral glass prism at an angle of incidence '$i$'. The emergent ray just grazes along the adjacent face. The refractive index of the prism is $\sqrt{2}$. The value of the angle of incidence at the first face is $(\sin 45^{\circ} = 1/\sqrt{2}, \sin 90^{\circ} = 1)$.

The refractive index of the material of a glass prism is $\sqrt{3}$. If the angle of minimum deviation is equal to the angle of the prism,then the angle of the prism is:
$\left(\cos 30^{\circ} = \frac{\sqrt{3}}{2} = \sin 60^{\circ}, \sin 30^{\circ} = \frac{1}{2} = \cos 60^{\circ}\right)$ (in $^{\circ}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo