A person has a near point of 50 , cm and a fa | vedclass

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

Question

(B) For $(i)$ reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at their actual near point $(v = -5

Vedclass · Accepted Answer

$2 \, D$ and $-0.33 \, D$

Vedclass · Answer

(B) For $(i)$ reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at their actual near point $(v = -50 \, cm)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50} \, cm^{-1}$.Power $P = \frac{100}{f(cm)} = \frac{100}{50} = +2 \, D$.For $(ii)$ seeing distant stars,the person needs to see an object at infinity $(u = \infty)$ as if it were at their far point $(v = -3 \, m)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-3} - \frac{1}{\infty} = -\frac{1}{3} \, m^{-1}$.Power $P = \frac{1}{f(m)} = -\frac{1}{3} \approx -0.33 \, D$.

$A$ person has a near point of $50 \, cm$ and a far point of $3 \, m$. What power of lens is required to $(i)$ read and $(ii)$ see distant stars?

Similar Questions

$A$ presbyopic patient has a near point of $30 \ cm$ and a far point of $40 \ cm$. The dioptric power of the corrective lens required for seeing distant objects is: (in $D$)

The sensation of vision in the retina is carried to the brain by

If the far point of a short-sighted person is $400 \ cm$, then the power of the lens required to enable him to see very distant objects clearly is

Calculate the focal length of a reading glass of a person if his distance of distinct vision is $ 75 \ cm $. (in $cm$)

$A$ person cannot see objects beyond $3 \, m$. What should be the focal length of the lens required to see objects at $12 \, m$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module