
Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water ${{\mu }_{w}}=\dfrac{4}{3}$, then the image formed from the mirror is at a distance of
Answer
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Hint:Here two convex glass pieces are joined together and one side is painted so that there could be no refraction at all. Since, the glass pieces are similar so the radius of curvature of the two sides must be equal. We are given that the second medium is now water between the two pieces, so we can make use of the lens maker formula to find the answer.
Formula used:
Thin lens formula: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Complete step by step answer:
When there is air between the two glass, the system behaves as a convex mirror. Upon introduction of water, the system will act as a convex mirror. The image distance and the object distance both are 20 cm, thus
Using thin lens formula: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{20}+\dfrac{1}{20}$
$\Rightarrow \dfrac{1}{f}=\dfrac{2}{20}$
$\Rightarrow f=10cm$
Also, $R=2f=20cm$
Upon introduction of water, the system will behave like a convex lens of water polished at one of the surfaces. This will act as concave mirror of focal length given by:
$\Rightarrow \dfrac{1}{f}=2\times 2\dfrac{(\mu -1)}{R}+\dfrac{2}{R}$
$\Rightarrow \dfrac{1}{f}=\dfrac{4\mu -2}{R}$
$\Rightarrow \dfrac{1}{f}=\dfrac{4\times \dfrac{3}{2}-2}{20}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{6}$
Now, using mirror formula:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Image is formed at the same place, thus
$\Rightarrow \dfrac{1}{u}+\dfrac{1}{u}=\dfrac{1}{6}$\[\]
$\Rightarrow \dfrac{2}{u}=\dfrac{1}{6} \\
\therefore u=12cm \\$
So, then the image formed from the mirror is at a distance of 12 cm.
Note:Convex lens is a converging lens and its focal length is positive. Concave lens is a diverging lens and its focal length is negative. While doing problems in optics, we have to keep in mind the sign convention. For convex lenses take the focal length as positive and for concave lenses take the focal length as negative. Convex lens converges parallel rays of light while a concave lens diverges parallel rays of light.
Formula used:
Thin lens formula: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Complete step by step answer:
When there is air between the two glass, the system behaves as a convex mirror. Upon introduction of water, the system will act as a convex mirror. The image distance and the object distance both are 20 cm, thus
Using thin lens formula: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{20}+\dfrac{1}{20}$
$\Rightarrow \dfrac{1}{f}=\dfrac{2}{20}$
$\Rightarrow f=10cm$
Also, $R=2f=20cm$
Upon introduction of water, the system will behave like a convex lens of water polished at one of the surfaces. This will act as concave mirror of focal length given by:
$\Rightarrow \dfrac{1}{f}=2\times 2\dfrac{(\mu -1)}{R}+\dfrac{2}{R}$
$\Rightarrow \dfrac{1}{f}=\dfrac{4\mu -2}{R}$
$\Rightarrow \dfrac{1}{f}=\dfrac{4\times \dfrac{3}{2}-2}{20}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{6}$
Now, using mirror formula:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Image is formed at the same place, thus
$\Rightarrow \dfrac{1}{u}+\dfrac{1}{u}=\dfrac{1}{6}$\[\]
$\Rightarrow \dfrac{2}{u}=\dfrac{1}{6} \\
\therefore u=12cm \\$
So, then the image formed from the mirror is at a distance of 12 cm.
Note:Convex lens is a converging lens and its focal length is positive. Concave lens is a diverging lens and its focal length is negative. While doing problems in optics, we have to keep in mind the sign convention. For convex lenses take the focal length as positive and for concave lenses take the focal length as negative. Convex lens converges parallel rays of light while a concave lens diverges parallel rays of light.
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