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Correct Answer: A
Solution :
For improving near point, convex lens is required and for this convex lens \[u=-\,25\ cm,\] \[v=-\,75\ cm\] \[\therefore \frac{1}{f}=\frac{1}{-75}-\frac{1}{-25}\Rightarrow f=\frac{75}{2}cm\] So power \[P=\frac{100}{f}=\frac{100}{75/2}=+\frac{8}{3}D\]
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Page 2
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Correct Answer: B
Solution :
In short sightedness, the focal length of eye lens decreases, so image is formed before retina.
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Page 3
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Correct Answer: D
Solution :
The image of object at infinity should be formed at 100 cm from the eye \[\frac{1}{f}=\frac{1}{\infty }-\frac{1}{100}=-\frac{1}{100}\] So the power \[=\frac{-100}{100}=-\ 1\]D (Distance is given in cm but \[P=\frac{1}{f}\] in metres)
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Page 4
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Correct Answer: B
Solution :
For improving far point, concave lens is required and for this concave lens \[u=\infty ,\ \ v=-\ 30\ cm\] So \[\frac{1}{f}=\frac{1}{-30}-\frac{1}{\infty }\Rightarrow f=-\ 30\ cm\] for near point \[\frac{1}{-30}=\frac{1}{-15}-\frac{1}{u}\Rightarrow u=-\ 30\ cm\]
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Page 5
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Correct Answer: C
Solution :
For myopic eye f = ? (defected far point) \[\Rightarrow f=-\,40\,cm\] Þ \[P=\frac{100}{-40}=-\,2.5\,D\]
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Page 6
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Correct Answer: C
Solution :
For lens u = want?s to see \[=-\,60\,cm\] v = can see \[=-\,10\,cm\] \[\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\]\[\Rightarrow \frac{1}{f}=\frac{1}{-10}-\frac{1}{(-60)}\]\[\Rightarrow f=-\,12\,cm\]
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Page 7
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Correct Answer: B
Solution :
Focal length = ? (Detected far point)
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Page 8
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Correct Answer: C
Solution :
In this case, for seeing distant objects the far point is 40 cm. Hence the required focal length is \[f=-\ d\](distance of far point) = ? 40 cm Power \[P=\frac{100}{f}cm=\frac{100}{-40}=-\ 2.5\ D\]
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Page 9
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Correct Answer: B
Solution :
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Page 10
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Correct Answer: A
Solution :
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Page 11
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Correct Answer: A
Solution :
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Page 12
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Correct Answer: A
Solution :
For viewing far objects, concave lenses are used and for concave lens u = wants to see \[=-\,60\,cm\]; v = can see\[=-\,15\,cm\] so from \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\]\[\Rightarrow f=-\,20\,cm\].
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Page 13
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Correct Answer: D
Solution :
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Page 14
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Correct Answer: A
Solution :
In short sightedness, the focal length of eye lens decreases and so the power of eye lens increases.
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Page 15
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Correct Answer: D
Solution :
Colour blindness is a genetic disease and still cannot be cured.
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Page 16
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Correct Answer: C
Solution :
Convexity to lens changes by the pressure applied by ciliary muscles.
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Page 17
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Correct Answer: B
Solution :
\[f=-\ d=-\ 100\ cm=-\ 1\ m\] \[\therefore P=\frac{1}{f}=\frac{1}{-1}=-\ 1\ D\]
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Page 18
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Correct Answer: C
Solution :
For correcting myopia, concave lens is used and for lens. u = wants to see \[=-\,50\,cm\] v = can see \[=-\,25\,cm\] From \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\]\[\Rightarrow \frac{1}{f}=\frac{1}{-25}-\frac{1}{(-50)}\]\[\Rightarrow f=-\,50\,cm\] So power \[P=\frac{100}{f}=\frac{100}{-50}=-\,2\,D\]
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Page 19
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Correct Answer: C
Solution :
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Page 20
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Correct Answer: C
Solution :
\[f=-\ d=-\ 60\ cm\]
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Page 21
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Correct Answer: B
Solution :
For correcting the near point, required focal length \[f=\frac{50\times 25}{(50-25)}=50\,cm\] So power \[P=\frac{100}{50}=+\,2\,D\] For correcting the far point, required focal length \[f=-\,(defected\ far\ point)=-\,3\,m\] \[\therefore P=-\frac{1}{3}D=-\,0.33\,D\]
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Page 22
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Correct Answer: B
Solution :
Negative power is given, so defect of eye is nearsigntedness Also defected far point \[=-f=-\frac{1}{p}=-\frac{100}{(-2.5)}=40\,cm\]
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Page 23
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Correct Answer: A
Solution :
In myopia, eye ball may be elongated so, light rays focussed before the retina.
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Page 24
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Correct Answer: C
Solution :
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Page 25
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Correct Answer: D
Solution :
\[P=\frac{1}{f}=\frac{1}{-(defected\ far\ point)}=-\frac{1}{2}\]\[=-\,0.5\,D\]
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Page 26
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Correct Answer: A
Solution :
Resolving limit of eye is one minute (1').
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