Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A presbyopic patient is having a near point as $30cm$ and far point as $40cm$. The dioptric power for the corrective lens in order to see the distant objects will be given as,\begin{align} & A.40D \\ & B.4D \\ & C.-2.5D \\ & D.0.25D \\ \end{align}

Last updated date: 20th Jun 2024
Total views: 346.5k
Views today: 3.46k
Verified
346.5k+ views
Hint: First of all the focal length of the lens should be found. As the far point is mentioned in order to see the distant objects, the focal length of the lens will be the same as the far point. Then the power of the lens is to be calculated. The power of the lens can be reciprocal of the focal length of the lens used.

In this situation, which has been mentioned in the question, the far point for seeing the object is given as,
$D=40cm$
As the far point is mentioned in order to see the distant objects, the focal length of the lens will be the same as the far point. Only the direction as per the sign convention will be different. Therefore the required focal length will be given as,
$f=-D$
Substituting the values in it will give,
$f=-40cm$
The power of a lens is given by the reciprocal of the focal length. Therefore the equation of the power can be written as,
$P=\dfrac{1}{f}$
Substituting the value of focal length in it,
$P=\dfrac{1}{-40}=-2.5D$
Therefore the question has been answered.

So, the correct answer is “Option C”.

Note: As the age of the people increases, the eye lens becomes harder and less elastic, and it makes it more difficult for the eye to focus on nearer objects. The presbyopia has been corrected with the use of bifocal eyeglasses for more than a century. Nowadays there are so many other ways to correct presbyopia using eyeglasses, contact lenses and also by surgery.