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STD 12 - 9. Ray optics abd optical instruments — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 9. Ray optics abd optical instruments4 Marks
Question
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

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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