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

For the myopic eye,the defect is cured by

$A$ person has a near point at $60\; cm$. The focal length of the spectacle lenses required to read at $22\; cm$,with the glasses separated by $2\; cm$ from the eyes,is: (in $; cm$)

$A$ man who cannot see clearly beyond $5 \, m$ wants to see stars clearly. He should use a lens of focal length ....... $m$.

$A$ 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 $, D$)

An imaginary line joining the optical centre of the eye lens and the yellow point is called as

Vedclass Test Series

Exam Paper Generator

Online Exam Module