A point object O is placed on the principal axis of a convex lens of focal length f=20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm . An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is. (assume paraxial approximation to be valid here)

Question

A point object O is placed on the principal axis of a convex lens of focal length f=20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm . An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is. (assume paraxial approximation to be valid here) (A) 0 (B) 2.5 cm (C) 5 cm (D) 10 cm

Tardigrade Admin · Accepted Answer

(1/ textv) = (1/f ) + (1/ textu) = (1/2 0) - (1/4 0) = (1/4 0) v = 40 cm Solution

To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan θ from axis to see the image i.e., 2 0 ((5/4 0)) = text2.5 cm Solution

0

$2.5 c m$

$5 c m$

$10 c m$

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$= \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$
v = 40 cm

To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan $θ$ from axis to see the image i.e., $20 (\frac{5}{40}) = 2.5 c m$

0

2.5cm

5cm

10cm

v1​=f1​+u1​ =201​−401​=401​ v = 40 cm To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan θ from axis to see the image i.e., 20(405​)=2.5cm

$2.5 c m$

$5 c m$

$10 c m$

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$= \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$
v = 40 cm

To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan $θ$ from axis to see the image i.e., $20 (\frac{5}{40}) = 2.5 c m$