The focal length \( f \) of a lens in a medium is given by the lens maker's formula:\[ \frac{1}{f} = (n_{\text{lens}} - n_{\text{medium}}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For a double convex lens, \( R_1 = 17 \, \text{cm} \) and \( R_2 = -17 \, \text{cm} \). The refractive index of glass \( n_{\text{lens}} = 1.5 \) and water \( n_{\text{medium}} = 1.33 \). Substituting these values: \[ \frac{1}{f} = (1.5 - 1.33) \left( \frac{1}{17} - \frac{1}{-17} \right) \] \[ \frac{1}{f} = 0.17 \left( \frac{2}{17} \right) = 0.17 \times \frac{2}{17} = 0.02 \, \text{cm}^{-1} \] \[ f = \frac{1}{0.02} = 50 \, \text{cm} \] Thus, the focal length of the lens in water is \( 50 \, \text{cm} \).
Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.
“One of these days you’re going to talk yourself into a load of trouble,” her father said aggressively. What do you learn about Sophie’s father from these lines? (Going Places)
Write a letter to the editor of a local newspaper expressing your concerns about the increasing “Pollution levels in your city”. You are an environmentalist, Radha/Rakesh, 46, Peak Colony, Haranagar. You may use the following cues along with your own ideas: 