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 who can see things most clearly at a distance of $10 \, cm$ requires spectacles to enable him to see clearly things at a distance of $30 \, cm$. What should be the focal length of the spectacles?

If a human had only one eye,then:

For a short-sighted person,the image is formed at

The retina of the eye acts like the ........ of a camera.

$A$ person wears glasses of power $-2.5 \, D$. The defect of the eye and the far point of the person without the glasses are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module