A person wears glasses of power -2.0 , D. The | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The power of the lens is $P = -2.0 \, D$. Since the power is negative,the lens used is a concave lens,which is used to correct myopia (nearsighted

Vedclass · Accepted Answer

Nearsighted,$50 \, cm$

Vedclass · Answer

(A) The power of the lens is $P = -2.0 \, D$. Since the power is negative,the lens used is a concave lens,which is used to correct myopia (nearsightedness).For a myopic eye,the far point is at a finite distance $x$ instead of infinity.The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.For a person to see an object at infinity $(u = \infty)$,the image must be formed at their far point $(v = -x)$.Since $P = \frac{1}{f}$,we have $P = \frac{1}{v} - \frac{1}{u} = \frac{1}{-x} - \frac{1}{\infty} = -\frac{1}{x}$.Thus,$x = -\frac{1}{P} = -\frac{1}{-2.0} = 0.5 \, m = 50 \, cm$.Therefore,the person is nearsighted and their far point is $50 \, cm$.

$A$ person wears glasses of power $-2.0 \, D$. The defect of the eye and the far point of the person without the glasses will be

Similar Questions

Hypermetropia is a:

The image formed on the retina is:

The near point and far point of a person are $40\,cm$ and $250\,cm$ respectively. Determine the power of the lens he/she should use while reading a book kept at a distance of $25\,cm$ from the eye $..........D$

When objects at different distances are seen by the eye,which of the following remains constant?

The least distance of distinct vision for a farsighted person is $1 \ m$. The optical power of the lens of his spectacles,which effectively reduces his least distance of distinct vision $(LDDV)$ to $25 \ cm$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module