MCQ
What will be the minimum angle of incidence such that the total internal reflection takes place from both the surface?.......$^o$


- A$30$
- B$45$
- ✓$60$
- D$75$

$\sin c=\frac{1}{\mu}=\frac{1}{\sqrt{2}}$
$c=45^{\circ}$
then after $T.I.R.$
$\sin c=\frac{\sqrt{3}}{2} \Rightarrow c=60^{\circ}$
minimum incidence angle $=60^{\circ}$
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($A$) The magnitude of induced emf in the wire is $\left(\frac{\mu_0}{\pi}\right)$ volt
($B$) If the loop is rotated at a constant angular speed about the wire, an additional emf of $\left(\frac{\mu_0}{\pi}\right)$ volt is induced in the wire
($C$) The induced current in the wire is in opposite direction to the current along the hypotenuse
($D$) There is a repulsive force between the wire and the loop