Question

An object of size 1.5cm is placed at a distance of 15cm from a convex lens.

An image is formed at 30cm from the lens. Calculate the
(i) size of the image and its nature
(ii) power of the lens​

Answer 1

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1. size of image is 3cm
2. nature of image is virtual and enlarged
3. power of lens is +0.3 dioptre

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Here,

Object distance , u = -15 cm ( it is to the left of lens)

image distance, v = -30cm ( it is to the left of lens)

object size = 1.5cm

symbols which we have to use :-

• object distance = u
• image distance = v
• object size = height of object = h2
• image size = height of image = h1
• focal length = f
• power of lens = p

we know that,

$m = \frac{v }{u}$

$\implies \: m = \frac{ - 30}{ - 15} \\ \\ \implies \: m = + 2$

since, the value of magnification is more than 1 ( it is 2 times more than image )

therefore the image is large than the object ( as given in question 15 < 30)

we know that..

$m = \frac{h2}{h1} \\ \\ \implies \: 2 = \frac{h2}{1.5} \: \: \: \: \: \: \: \: \\ \\ \implies \: h2 = 2 \times 1.5 \\ \\ \implies \: h2 = 3cm \: \: \: \: \: \:$

so size of image is 3cm

and nature of image is virtual ( magnification is positive)

we know that..

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ \implies \: \frac{1}{f} = \frac{1}{ - 30} - \frac{1}{ - 15} \\ \\ \implies \: \frac{1}{f} = \frac{ - 1 + 2}{30} \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \frac{1}{f} = \frac{ 1}{30} \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \: f = 30cm \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so focal length of the lens is 30cm

and we know that,

$p = \frac{1}{f(in \: metres)} \\ \\ \implies \: p = \frac{1}{0.3m}$

so power of lense is 0.3 dioptre