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 \( {|A_2|} = {3^\circ} \).
Given below are two statements:
Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$).
Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below:
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below: