Question

An object 2.5 cm long is placed on the axis of a concave mirror of 30 cm radius of curvature at a distance of 10 cm from it. Find the position, size and nature of the image formed.​

Answer 1

Answer:

✡30cm , magnified, virtual and erect. ☑

Explanation:

Given ⤵

➡Distance u=−10cm

➡Height h=2.5cm

➡Radius of curvature R=30cm

⚫We know that

$−f= \frac{R}{2}\\ −f= \frac{30}{2}\\ f=−15cm$

Now, using mirror’s formula

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{-15}= \frac{1}{v}+{1}{-10}\\ \frac{1}{v}=\frac{-1}{15}+\frac{1}{10}\\v=30cm$

⚫Now, the magnification is

$m=\frac{-v}{u}\\m=\frac{-30}{-10}\\m=3$

We know that,

$m= \frac{h'}{h}\\3=\frac{h'}{2.5}\\h=7.5cm$

Hence, the size of the image is 7.5 cm.

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