Question

Convex mirror of focal length f forms an image which 1/n times the object.The distance of object form the mirror is ​

Answer 1

Given:

Focal length of convex mirror = f

Size\height of image w.r.t size\height of object i.e. Magnification $\sf \bigg(\dfrac{h_i}{h_o} \bigg)$ = $\sf \dfrac{1}{n}$

To Find:

Object distance (u)

Answer:

As convex mirror always forms a virtual image which is diminished in size. So,

$\rm \implies \dfrac{h_i}{h_o} = \dfrac{1}{n} \\ \\ \rm \implies -\dfrac{v}{u} = \dfrac{1}{n} \\ \\ \rm \implies v = -\dfrac{u}{n}$

i.e. for real object, object distance (u) will be negative and from the relation image distance (v) will be positive.

By Mirror Equation,

$\bf \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ \rm \implies \dfrac{1}{ \bigg( - \dfrac{u}{n} \bigg)} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ \rm \implies - \dfrac{n}{u} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ \rm \implies \dfrac{1 - n}{u} = \dfrac{1}{f} \\ \\ \rm \implies u = f(1 - n)$

$\therefore$ $\boxed{\mathfrak{Object \ distance \ (u) = -f(1-n) }}$

Note: Negative sign in object distance indicates that the object is real and kept in front of the mirror.

Answer 2

Answer:

GIVEN:

• Image height (h') = 1/n (Object height [h])

TO FIND:

• The distance of object from the mirror.

SOLUTION:

We know that:

$\sf{\red{m=\dfrac{h}{h'} =\dfrac{v}{u} }}$_______(1)

And It is given that:

$\sf h'=\left(\dfrac{1}{n} \times h \right)$____(2)

Substitute (1) in (2).

$\sf{\red{\dfrac{\not{h}}{\left(\dfrac{1}{n} \times \not{h} \right)} =\dfrac{v}{u} }}$

$\sf v=-(\dfrac{1}{n} \times u)$_____(3)

• The reason RHS is negative is because object distance (u) is negative.

Now, use the mirror formula:

$\sf \dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$____(4)

Substitute (3) in (4).

$\sf \dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{- \left(\dfrac{1}{n} \times u \right)}$

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

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

• Cross multiply.

$\sf u= f(n-1)$

The distance of object form the mirror is f(n-1) units.