Please be correct with this answer need quick help on hard a

Please be correct with this answer, need quick help on hard advanced physics question.

A nearsighted eye has an effective Ins to retina distance of 2.1 cm. This eye has a near point of 5 tin and a farpoint of 12 cm. What is the effective focal length of this eye when relaxed. What is the smallest possible focal length of this eye?

Solution

Given that lens to retina distance = 2.1cm

near point =5cm ; far point =12 cm

A) when eye is relaxed rays coming from far point will be focused on the ratina

there fore effective focal length will be given by

1/f = 1/u + 1/v

1/f = 1/12 + 1/2.1

f= (12*2.1)/(12+2.1)

f=1.79 cm

focal lenth = 1.79 cm

B)

The smallest possible focal length will be when rays from near point will be focused on ratina,

Therefore focal length will be given by

1/f= 1/5 + 1/2.1

f= (5*2.1)/(5+2.1)

f=1.48 cm

Smallest focal length = 1.48 cm