Question

an object is placed perpendicular to the principal axis of a convex lens of focal length 20cm. The distance of the object from the lens is 30 cm. Find (1) the position (2) magnification and (3)the nature of the image formed
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Answer 1

Given :

• Focal length (f) = 20 cm.

• Object Distance (u) = 30 cm.

To find :

• Image distance (v)
• Magnification (m)
• Nature of the image.

Solution :

Image distance :

We know the lens formula i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀1/f = 1/v - 1/u

Where :

• f = Focal length
• v = Final Velocity
• u = Image distance

By sign convention , we know that the object distance is always taken as negative in a convex mirror,i.e,

• u = (-30)

Now using the lens formula and substituting the values in it, we get :

==> 1/f = 1/v - 1/u

==> 1/20 = 1/v - 1/(-30)

==> 1/20 = 1/v + 1/30

==> 1/20 - 1/30 = 1/v

==> (30 - 20)/60 = 1/v

==> 10/60 = 1/v

==> 1/6 = 1/v

By cross-multiplication , we get :

==> 1 × v = 1 × 6

==> v = 6

∴ v = 6 cm.

Hence the object distance is 6 cm.

Magnification :

We know the formula for Magnification i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀m = - v/u

Where :

• m = Magnification
• v = Image distance
• u = Object Distance

Now using the formula for magnification and substituting the values in it, we get :

==> m = - v/u

==> m = - 6/(-30)

==> m = - (1/-5)

==> m = 1/5

==> m = 0.2

∴ m = 0.2

Hence the Magnification of the image is 0.2.

Nature of the image : The image is virtual and erect.

Answer 2

Answer:

Answer of kitty is wrong

Explanation:

value of v=60

it will form real and inverted image

as magnification is -2