A convex lens is put \(10\, cm\) from a light source and it makes a sharp image on a screen, kept \(10\, cm\) from the lens. Now a glass block (refractive index \(1.5\)) of \(1.5\, cm\) thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance \(d\). Then \(d\) is

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Answer & Solution

Correct Answer

(D) \(0.55\, cm\) away from the lens

Detailed explanation

If \(u=-10 \,\mathrm{cm}\)

\(v=+10\, \mathrm{cm}\)

\(\Rightarrow \mathrm{f}=5\,\mathrm{cm}\)

Glass plate shift \(=\mathrm{t}\left(1-\frac{1}{\mu}\right)=1.5\left(1-\frac{2}{3}\right)=0.5\,\mathrm{cm}\)

So, new \(u=10-0.5=9.5\,\mathrm{cm}\)

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}-\frac{1}{-9.5}=\frac{1}{5}\)

After solving we get,

\(v=\frac{47.5}{4.5}\). Hence, shift \(\frac{47.5}{4.5}-10=\left(\frac{2.5}{4.5}\right)=0.55\, \mathrm{cm}\)

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