Using the lens maker’s formula:
\[\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{air}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
Given:
\[f = 20 \, \text{cm}, \quad R_1 = 15 \, \text{cm}, \quad R_2 = -30 \, \text{cm}\]
Substitute into the formula:
\[\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)\]
Simplify the expression:
\[\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)\]
\[\Rightarrow \mu - 1 = \frac{1}{2}\]
\[\Rightarrow \mu = 1 + \frac{1}{2} = \frac{3}{2} = 1.5\]
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).