A student suffering from myopia is not able to see distinctly objects placed beyond 5m list two possible due to which this defect of vision may have arisen with the help of ray diagram explain first why a student is unable to see a distinctively the objects placed beyond 5m from his eyes secondly which type of corrective lens to restore proper vision how this defect can be corrected by the use of this lens
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There are some possible reason due to this defect of vision may have arisen :
(a) elongation of eyeball
(b) excessive curvature of the lens.
(i) student unable to see distinctly the objects placed beyond 5m from his eyes because student suffering with myopia . A myopic eye can't see distinct object clearly but able to see nearby object clearly.
(ii) concave lens of suitable focal length is used to restore proper vision. After using concave lens , image will form at retina. While before it , image was formed near the retina .
As shown in figure it is clear how the lens correct this type of defect.
b. Focal length of concave lens is negative
So, f = -5m
Now, Power = 1/f = -1/5 = -0.2D
Hence, power is -0.2 diopter
Hope my answer helps you ☺✌✌
There are some possible reason due to this defect of vision may have arisen :
(a) elongation of eyeball
(b) excessive curvature of the lens.
(i) student unable to see distinctly the objects placed beyond 5m from his eyes because student suffering with myopia . A myopic eye can't see distinct object clearly but able to see nearby object clearly.
(ii) concave lens of suitable focal length is used to restore proper vision. After using concave lens , image will form at retina. While before it , image was formed near the retina .
As shown in figure it is clear how the lens correct this type of defect.
b. Focal length of concave lens is negative
So, f = -5m
Now, Power = 1/f = -1/5 = -0.2D
Hence, power is -0.2 diopter
Hope my answer helps you ☺✌✌
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niya1694:
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2
Answer: the problem may be arrived due the following reason:
1) elongation of the eye balls
2) increase in curvature
Focal length= -5 mtr
Power=1÷f
=1÷5
=0.2 D
Thanks for asking
Hope it helps you
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