JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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
- A \(1.1\, cm\) away fro the lens
- B \(0\)
- C \(0.55\, cm\) towards the lens
- D \(0.55\, cm\) away from the lens
Answer & Solution
Correct Answer
(D) \(0.55\, cm\) away from the lens
Step-by-step Solution
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 \(u=10-0.5=9.5\,\mathrm{cm}\) \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)…
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