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 \( {|A_2|} = {3^\circ} \).
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.