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Answer :
The power P of a lens of focal length f is given by the relationP= \(\frac{1}{f} \)
(i) Power of the lens used for correcting distant vision = - 5.5 D
Focal length of the required lens,
f= \(\frac{1}{Pf} =\frac{1}{-5.5} = -0.181 m \)
The focal length of the lens for correcting distant vision is - 0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f= \(\frac{1}{P} \)
f= \(\frac{1}{1.5} \) = +0.667 m
The focal length of the lens for correcting near vision is 0.667 m.