The far point of a myopic person is $150 \ cm$ in front of the eye. Calculate the focal length and the power of a lens required to enable him to see distant objects clearly.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(D) Given: The object distance $u = \infty$ (for distant objects) and the image distance $v = -150 \ cm = -1.5 \ m$ (since the image must be formed at the person's far point).
Using the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-1.5} - \frac{1}{\infty}$.
Since $\frac{1}{\infty} = 0$, we get $\frac{1}{f} = -\frac{1}{1.5}$.
Therefore, the focal length $f = -1.5 \ m$.
The power of the lens $P$ is given by $P = \frac{1}{f(\text{in meters})}$.
$P = \frac{1}{-1.5} = -0.67 \ D$.
Thus, the required focal length is $-1.5 \ m$ and the power is $-0.67 \ D$.

Explore More

Similar Questions

How can you identify the type of vision defect a person is suffering from by physically touching their spectacles?

Which color is produced by mixing blue and red?

$(a)$ What is "power of accommodation of the eye"?
$(b)$ What happens to the image distance when the object being viewed is moved away from the eye?

Which of the following relationships regarding the wavelengths $\lambda$ of different colors of light is correct?

Identify the phenomena of light involved in the formation of a rainbow.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo