Question

An object and its real image are located at distances
25 cm and 40 cm respectively from the two
principal focii of a convex lens. The linear
magnification of the image is near to
(1) + 1.3
(2) -1.3
(3) + 1.8
(4) -1.8
​

Answer 1

answer : option (2) -1.3

explanation : let focal length of convex lens is f cm.

so, object distance from the pole of lens, u = -(25 + f) cm

image distance from the pole of lens, v = (40 + f) cm

now using lens maker formula, 1/v - 1/u = 1/f

⇒1/(40 + f) - 1/-(25 + f) = 1/f

⇒1/(40 + f) + 1/(25 + f) = 1/f

⇒(25 + f + 40 + f)/(40 + f)(25 + f) = 1/f

⇒65f + 2f² = 1000 + 40f + 25f + f²

⇒f² = 1000

⇒f = 31.6 cm

so, u = -(25 + 31.6) = -56.6 cm

v = (40 + 31.6) = 76.6 cm

now magnification = v/u = (76.6)/(-56.6)

≈ -1.3

hence, option (2) is correct choice.

Answer 2

Answer:

so simpoooool

use newton's law =  focal  length =√x1x2

f = √40*25 =√1000  =10 √10  =10*3.16  =31.6

now add the 'U' and 'V" values because all distance are measured from optical center

"V" = 31.6+40 =71.6

"U" = -(25 +31.6) = -56.6

so the magnification is v/u let  put it up

71.6/-56.6  =  -1.3

Explanation: