A student can distinctly see objects up to a | vedclass

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

Question

(D) The student suffers from myopia (nearsightedness). To see objects at infinity (or a far distance),the lens must form a virtual image of the object

Vedclass · Accepted Answer

$ - 15.8 \, cm, - 6.3 \, D$

Vedclass · Answer

(D) The student suffers from myopia (nearsightedness). To see objects at infinity (or a far distance),the lens must form a virtual image of the object at the student's far point.Given: Far point $v = -15 \, cm$,Object distance $u = -300 \, cm$ $(3 \, m = 300 \, cm)$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-15} - \frac{1}{-300} = \frac{-20 + 1}{300} = \frac{-19}{300}$.Therefore,$f = -\frac{300}{19} \approx -15.8 \, cm$.The power of the lens is given by $P = \frac{100}{f (cm)} = \frac{100}{-15.789} \approx -6.33 \, D$.Thus,the focal length is $-15.8 \, cm$ and the power is $-6.3 \, D$.

$A$ student can distinctly see objects up to a distance of $15 \, cm$. He wants to see the blackboard at a distance of $3 \, m$. The focal length and power of the lens used,respectively,will be:

Similar Questions

$A$ person wears normal spectacles in which the distance between the glasses and the eyes is approximately $2 \, cm$. The power required is $-5 \, D$. If he wears contact lenses,then the required power is ......... $D$.

$A$ person wears glasses of power $-2.5 \, D$. What is the far point of the person without glasses in $cm$?

For a normal eye,the least distance of distinct vision is......$m$.

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:

The impact of an image on the retina remains for......$sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module