Question 221 of 295
For an object placed at a distance from a lens, a sharp focused image is observed on a screen placed at a distance from the lens. A glass plate of refractive index and thickness is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
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Answer: (B)
Given:
object using the lens formula:
\(\begin{aligned}& \frac{1}{f}=\frac{1}{v \prime}-\frac{1}{w^{\prime}} \\\Rightarrow & \frac{21}{240} \mathrm{~cm}=\frac{3}{35} \mathrm{~cm}-\frac{1}{u^{\prime}} \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{3}{35} \mathrm{~cm}-\frac{21}{240} \mathrm{~cm} \\\Rightarrow & \frac{1}{u^{\prime}}=\frac{1}{5}\left[\frac{3}{7} \mathrm{~cm}-\frac{21}{48} \mathrm{~cm}\right] \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{1}{5}\left[\frac{144-147}{336}\right] \\\Rightarrow u^{\prime} & =-560 \mathrm{~cm}=-5.6 \mathrm{~m}\end{aligned}\)
Hence, the object should be at \(5.6 \mathrm{~m}\) in front of the lens.
Hence, the oblect should be moved by 5.6-24= \(3.2 \mathrm{~cm}\).
object using the lens formula:
\(\begin{aligned}& \frac{1}{f}=\frac{1}{v \prime}-\frac{1}{w^{\prime}} \\\Rightarrow & \frac{21}{240} \mathrm{~cm}=\frac{3}{35} \mathrm{~cm}-\frac{1}{u^{\prime}} \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{3}{35} \mathrm{~cm}-\frac{21}{240} \mathrm{~cm} \\\Rightarrow & \frac{1}{u^{\prime}}=\frac{1}{5}\left[\frac{3}{7} \mathrm{~cm}-\frac{21}{48} \mathrm{~cm}\right] \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{1}{5}\left[\frac{144-147}{336}\right] \\\Rightarrow u^{\prime} & =-560 \mathrm{~cm}=-5.6 \mathrm{~m}\end{aligned}\)
Hence, the object should be at \(5.6 \mathrm{~m}\) in front of the lens.
Hence, the oblect should be moved by 5.6-24= \(3.2 \mathrm{~cm}\).
