Given:
\( \mu_1 = 1, \quad \mu_2 = 1.5, \quad R = 20 \, \text{cm}, \quad \text{Object distance} = 100 \, \text{cm} \)
Step 1: Using the refraction formula
The relation between the refractive indices and distances is given by: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Step 2: Substituting the known values
\[ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20} \]
Step 3: Solving for \( v \)
\[ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} \]
\[ \frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100} \]
\[ \frac{1.5}{v} = \frac{5}{100} - \frac{1}{100} = \frac{4}{100} \]
\[ v = \frac{1.5 \times 100}{4} = 100 \, \text{cm} \]
Step 4: Finding the total distance
The total distance from the object is: \[ \text{Distance from object} = 100 + 100 = 200 \, \text{cm} \]
Final Answer:
\[ \boxed{200 \, \text{cm}} \]
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:
\[ 100 + 100 = 200 \, \text{cm from the object}. \]A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 