Given: - Radius of curvature \( R = 20 \, \text{cm} \), - Refractive indices: \( \mu_1 = 1 \) (air) and \( \mu_2 = 1.5 \) (medium).
Using the lens maker’s formula:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. \]Substitute \( u = -100 \, \text{cm} \):
\[ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}. \]Solving for \( v \):
\[ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}, \] \[ \frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100}, \] \[ v = 100 \, \text{cm}. \]Thus, the image is formed at a distance:
\[ 100 + 100 = 200 \, \text{cm from the object}. \]List-I | List-II | ||
P | If \(n = 2\) and \(\alpha = 180°\), then all the possible values of \(\theta_0\) will be | I | \(30\degree\) or \(0\degree\) |
Q | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be | II | \(60\degree\) or \(0\degree\) |
R | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be | III | \(45\degree\) or \( 0\degree\) |
S | If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of \(\alpha\) will be | IV | \(150\degree\) |
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Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32