Question

A needle 10 cm high is placed at 40 cm from a convex lens of focal length 15 cm find position nature and size of image also draw ray diagram​

Answer 1

Given:-

• Height of the object (needle) = 10 cm
• Object distance = 40 cm
• Focal length of convex lens = 15 cm

To Find:-

• Position of the image
• Nature of the image
• Size of the image

Note:-

• Refer to the attachment for the ray diagram.

Answer:-

We know,

According to the sign convention, we know, that the object distance is negative in all cases. In a convex lens, the focal length is positive since it is present on the right hand side of the lens. also the image distance is positive.

Therefore:-

• Object distance = -40 cm
• Focal length = 15 cm

Also we know,

Lens formula is as follows:-

$\sf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

Where:-

• f = focal length of the mirror
• v = image distance
• u = object distance

Putting the values in the formula,

$= \sf{\dfrac{1}{15} = \dfrac{1}{v} - \dfrac{1}{-40}}$

$= \sf{\dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{40}}$

$= \sf{\dfrac{1}{15} - \dfrac{1}{40} = \dfrac{1}{v}}$

$= \sf{\dfrac{8 - 3}{120} = \dfrac{1}{v}}$

$= \sf{\dfrac{5}{120} = \dfrac{1}{v}}$

$= \sf{\dfrac{120}{5} = v}$

$= \sf{v = 24}$

∴ The image distance (or position of the image) is 24 cm

Now,

We know,

Magnification of a lens is given by:-

$\sf{m = \dfrac{v}{u}}$

Therefore,

$= \sf{m = \dfrac{24}{-40}}$

$= \sf{m = \dfrac{24}{-40}}$

$= \sf{m = -0.6}$

∴ The nature of the image is real and inverted

Also for height of the image:-

$= \sf{m = \dfrac{v}{u} = \dfrac{h_i}{h_o}}$

Where,

• $\sf{h_i = Height\:of\:image}$
• $\sf{h_o = Height\:of\:image}$
• m = magnification

Note: Height of the object is always positive according to the sign convention.

So:-

$= \sf{-0.6 = \dfrac{h_i}{10}}$

$= \sf{-0.6 \times 10 = h_i}$

$= \sf{h_i = -6}$

∴ The height of the image is 6 cm.

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Additional Information:-

• The image formed by a convex lens is real and inverted if the magnification is negative.

• The image formed by the convex lens is virtual and erect if the magnification is positive.

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