Step 1: Relation in Prism
From the prism relation:
$$ r_1 + c = A $$
Rearranging for \( r_1 \):
$$ r_1 = 90^\circ - c \quad \text{...(1)} $$
Step 2: Expression for \( \cos c \)
We know:
$$ \sin c = \frac{1}{\mu} $$
Using trigonometric identity:
$$ \cos c = \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Step 3: Apply Snell's Law on Incidence Surface
Using Snell’s law at the first surface:
$$ \sin 30^\circ = \mu \sin (r_1) $$
Substituting \( r_1 = 90^\circ - c \):
$$ \frac{1}{2} = \mu \sin (90^\circ - c) $$
Since \( \sin (90^\circ - c) = \cos c \), we get:
$$ \frac{1}{2} = \mu \times \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Simplifying:
$$ \frac{1}{2} = \frac{\sqrt{\mu^2 - 1}}{1} $$
Step 4: Solve for \( \mu \)
Squaring both sides:
$$ \frac{1}{4} = \mu^2 - 1 $$
Rearranging:
$$ \mu^2 = \frac{5}{4} $$
Taking square root:
$$ \mu = \frac{\sqrt{5}}{2} $$
Conclusion
The refractive index \( \mu \) of the prism is \(\frac{\sqrt{5}}{2}\).
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :