Question

plz answer this question.....
an object of height 2 cm is placed at a distance 20 cm in front of a concave mirror of focal length 12 cm find the position,size,nature of the image ​

Answer 1

ANSWER:

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Size of object h1 = 2cm

Size of the image h2=?

Focal length =12cm

Object distance u = -20cm

Image distance v =?

Now,

using mirror formula:

$\\ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} ........(formula)$

Putting the values in the above formula.

$\\ = \frac{1}{ - 12} = \frac{1}{v} - \frac{1}{ - 20}$

$\\ \frac{1}{v} = \frac{ - 5 + 3}{60} ......(taking \: the \: lcm) \\ \\ \frac{1}{v} = \frac{ - 2}{60} \\ \\ v = - 30$

Now,we have the image distance v=-30cm

So,

now moving towards the magnification of the image.

$\\ magnification = \frac{h1}{h2} = \frac{ - v}{u} = \frac{30}{ - 20 } \\ \\ magnification = - 1.5$

Now we find the size of the image =h2=?

$h2 = - 1.5 \times 2 \\ \\ h2 = - 3cm \\ \\ h2 = 3cm$

Therefore the image formed will be real, inverted and magnified.

Answer 2

Answer:

The image will be formed 30 cm in front of the mirror.

The size of the image will be 3 cm.

The image will be real, inverted and highly magnified.

Explanation:

Referred to the attachments.