To solve this problem, we need to find the angle \( A_2 \) of prism \( P_2 \) such that the combination of prism \( P_1 \) with angle \( A_1 = 4^\circ \) and refractive index \( n_1 = 1.54 \) and prism \( P_2 \) with refractive index \( n_2 = 1.72 \) produces dispersion without deviation.
1. Recall the formula for deviation by a thin prism:
The deviation \( \delta \) caused by a thin prism with angle \( A \) and refractive index \( n \) is given by:
\( \delta = (n - 1)A \)
2. Apply the condition for zero net deviation:
For dispersion without deviation, the net deviation must be zero. Therefore, the sum of the deviations caused by the two prisms must be zero:
\( \delta_1 + \delta_2 = 0 \)
3. Substitute the expressions for \(\delta_1\) and \(\delta_2\):
Substitute \( \delta_1 = (n_1 - 1)A_1 \) and \( \delta_2 = (n_2 - 1)A_2 \) into the equation \( \delta_1 + \delta_2 = 0 \):
\( (n_1 - 1)A_1 + (n_2 - 1)A_2 = 0 \)
4. Substitute the given values:
Substitute the given values \( n_1 = 1.54 \), \( A_1 = 4^\circ \), and \( n_2 = 1.72 \) into the equation:
\( (1.54 - 1)(4) + (1.72 - 1)(A_2) = 0 \)
\( (0.54)(4) + (0.72)(A_2) = 0 \)
5. Solve for \(A_2\):
Solve for \( A_2 \):
\( 2.16 + 0.72 A_2 = 0 \)
\( 0.72 A_2 = -2.16 \)
\( A_2 = -\frac{2.16}{0.72} = -3 \)
6. Interpret the result:
The angle of the prism \( P_2 \) is \( A_2 = -3^\circ \). The negative sign indicates that the prism \( P_2 \) is placed in an inverted position relative to the prism \( P_1 \). Since we are asked for the magnitude of the angle, we take the absolute value.
Final Answer:
The angle of prism \( P_2 \) is \( {|A_2|} = {3^\circ} \).
Match List - I with List - II:
List - I:
(A) Electric field inside (distance \( r > 0 \) from center) of a uniformly charged spherical shell with surface charge density \( \sigma \), and radius \( R \).
(B) Electric field at distance \( r > 0 \) from a uniformly charged infinite plane sheet with surface charge density \( \sigma \).
(C) Electric field outside (distance \( r > 0 \) from center) of a uniformly charged spherical shell with surface charge density \( \sigma \), and radius \( R \).
(D) Electric field between two oppositely charged infinite plane parallel sheets with uniform surface charge density \( \sigma \).
List - II:
(I) \( \frac{\sigma}{\epsilon_0} \)
(II) \( \frac{\sigma}{2\epsilon_0} \)
(III) 0
(IV) \( \frac{\sigma}{\epsilon_0 r^2} \) Choose the correct answer from the options given below:
Consider the following statements:
A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface of a liquid.
B. As the temperature of liquid rises, the coefficient of viscosity increases.
C. As the temperature of gas increases, the coefficient of viscosity increases.
D. The onset of turbulence is determined by Reynolds number.
E. In a steady flow, two streamlines never intersect.
Choose the correct answer from the options given below: