Answer

Verified

97.8k+ views

**Hint**We are given here with the height of the object, object distance from the mirror and the radius of curvature of the mirror and are also told about the type of mirror. First, we will find the focal length of the mirror as it is defined as half of the radius of curvature. Then we will apply the mirror formula taking into account the sign convention. We will then use the formula for magnification in the mirror to complete the first question. Then using our findings, we will draw a suitable ray diagram.

**Formula Used**

$f = \dfrac{R}{2}$

Where, $f$ is the focal length of the mirror and $R$ is the radius of curvature of the mirror.

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

Where, $v$ is the image distance and $u$ is the object distance from the mirror.

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

Where, $m$ is the magnification by the mirror, ${h_i}$ is the height of the image and ${h_o}$ is the height of the object.

**Complete Step By Step Solution**

(i)

Here,

Given,

$R = 30cm$

Thus,

Focal length of the mirror, $f = \dfrac{{30}}{2}cm = 15cm$

But the mirror is a concave one.

Thus, the focal length is $f = - 15cm$

Also,

Given,

$u = 10cm$

But it is placed ahead of the mirror.

Thus,

$u = - 10cm$

Now,

We apply the mirror formula,

$\dfrac{1}{{ - 15}} = \dfrac{1}{v} + \dfrac{1}{{ - 10}}$

After further calculations, we get

$v = 30cm$

Thus, the image is formed $30cm$ in behind the mirror.

Now,

For the size and nature of the image, we apply the magnification formula.

$m = \dfrac{{ - 30}}{{ - 10}} = 3$

As the magnification is positive, thus the image formed is virtual and erect and since the value of the magnification is greater than one thus the image is magnified by three times.

Now,

$m = \dfrac{{{h_i}}}{{{h_o}}}$

Putting here the values,

$3 = \dfrac{{{h_i}}}{5} \Rightarrow {h_i} = 15cm$

The size of the image is $15cm$ in height.

(ii)

Here,

$AB$ is the object and \[{A^I}{B^I}\] is the image.

**Note**The mirror was concave thus we took the focal length to be negative. If the mirror was convex, then we should take the focal length as positive and then the whole calculation and the case would be different.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

Other Pages

Calculate CFSE of the following complex FeCN64 A 04Delta class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

The focal length of a thin biconvex lens is 20cm When class 12 physics JEE_Main

If two bulbs of 25W and 100W rated at 200V are connected class 12 physics JEE_Main

A ball of mass 05 Kg moving with a velocity of 2ms class 11 physics JEE_Main