A person uses a lens of power +3d to normalise his vision the focal length of the lens is

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Correct Answer: A

Solution :
Focal length of the lens \[f=\frac{100}{3}cm\] By lens formula \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\Rightarrow \frac{1}{+100/3}=\frac{1}{v}-\frac{1}{-25}\Rightarrow v=-100\ cm=-\ 1\ m\]

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Page 2

Correct Answer: D

Solution :

This is the defect of hypermetropia.

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Page 3

Correct Answer: A

Solution :
For large objects, large image is formed on retina.

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Page 4

Correct Answer: D

Solution :

\[v=-15cm,\] \[u=-\,300cm,\] From lens formula \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\Rightarrow \frac{1}{f}=\frac{1}{-15}-\frac{1}{-300}=\frac{-19}{300}\]\[\Rightarrow f=\frac{-\,300}{19}=-\,15.8\,cm\] and power \[P=\frac{100}{f}cm\]\[=\frac{-100\times 19}{300}\]= ? 6.33 D.

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Correct Answer: D

Solution :
Time of exposure \[\propto \frac{1}{{{(Aperture)}^{2}}}\]

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Page 6

Correct Answer: A

Solution :
Light gathering power µ Area of lens aperture or d2

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Page 7

Correct Answer: B

Solution :
Time of exposure \[\propto {{(f.\,number)}^{2}}\] \[\Rightarrow \frac{{{t}_{2}}}{{{t}_{1}}}={{\left( \frac{5.6}{2.8} \right)}^{2}}=4\] \[{{t}_{2}}=4\,{{t}_{1}}=4\times \frac{1}{200}=\frac{1}{50}\]sec = 0.02 sec.

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