A man can see only between 75 cm and 200 cm. the power of lens to correct the near point will be:

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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

• 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

• 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

• 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

• 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

• 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

• Correct Answer: B

  Solution :

             Focal length = ? (Detected far point)

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Page 8

• 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

• Correct Answer: B

  Solution :

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Page 10

• Correct Answer: A

  Solution :

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Page 11

• Correct Answer: A

  Solution :

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Page 12

• 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

• Correct Answer: D

  Solution :

    

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Page 14

• 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

• Correct Answer: D

  Solution :

             Colour blindness is a genetic disease and still cannot be cured.

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Page 16

• Correct Answer: C

  Solution :

             Convexity to lens changes by the pressure applied by ciliary muscles.

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Page 17

• 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

• 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

• Correct Answer: C

  Solution :

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Page 20

• Correct Answer: C

  Solution :

             \[f=-\ d=-\ 60\ cm\]

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Page 21

• 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

• 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

• Correct Answer: A

  Solution :

             In myopia, eye ball may be elongated so, light rays focussed before the retina.

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Page 24

• Correct Answer: C

  Solution :

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Page 25

• 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

• Correct Answer: A

  Solution :

             Resolving limit of eye is one minute (1').

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