#### PrincePhoenix

Gold Member

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**A small object of size 0.5 mm is placed close to a convex lens of focal length 5cm. The virtual image formed is at a distance of 25 cm from the lens. Find magnification.**

Data:

focal length, f = 5cm

distance of image , q = -25 cm (because image is virtual)

distance of object , p = ?

magnification, M = ?

Data:

focal length, f = 5cm

distance of image , q = -25 cm (because image is virtual)

distance of object , p = ?

magnification, M = ?

**1/f = 1/p + 1/q**

M = height of image/height of object = q/p

(ratio between height and distance of image and object)

M = height of image/height of object = q/p

(ratio between height and distance of image and object)

**Solution:**

First finding p,

1/f = 1/p + 1/q

1/p = 1/f - 1/q

1/p = 1/5 - (1/-25)

..................

1/p = 6/25

p = 25/6 = 4.16cm

Now putting value of 'p' in magnification formula,

M = height of image/height of object = q/p

(ratio between height and distance of image and object)

M = -25/4.16

M = -6

Can magnification be -ve? It is so in this problem. What have I done wrong?

First finding p,

1/f = 1/p + 1/q

1/p = 1/f - 1/q

1/p = 1/5 - (1/-25)

..................

1/p = 6/25

p = 25/6 = 4.16cm

Now putting value of 'p' in magnification formula,

M = height of image/height of object = q/p

(ratio between height and distance of image and object)

M = -25/4.16

M = -6

Can magnification be -ve? It is so in this problem. What have I done wrong?

2nd problem:

1-This part of a numerical problem as our teacher did it in classroom.

__Data:__

distance of object, p = 20cm

focal length of concave lens,f = -19 (Is focal length of concave lens really always -ve?)

distance of image, q = ?

__Solution:__

According to lens formula,

1/f = 1/p + 1/q

=> 1/q = 1/f - 1/p

1/q = 1/-19 - 1/20

1/q = -20-19/380

1/q = -39/380

q = -9.74cm

Why can't LCM be negative at line 4,5 of solution when one of the denominators is -ve? And is the -ve value of distance of image correct?