First slide

The optical instruments

Question

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

Moderate

A

Farsightedness, 40 cm
B

Nearsightedness, 40 cm
C

Astigmatism, 40 cm
D

Nearsightedness, 250 cm

Solution

Negative power is given, so defect of eye is nearsigntedness
$Also defected far point = - f = - \frac{1}{p} = - \frac{100}{(- 2 .5)} = 40 cm$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App