Question

Rohit placed a pencil perpendicular to principal Axes in front of a converging mirror of focal length 30 cm the image formed is twice the size of the pencil calculate distance of the object from the mirror

Answer 1

Answer:

Magnification=hiho=−vuMagnification=hiho=−vu

For real image

m=−vu=−2m=−vu=−2

v=2u

Now Using the mirror equation,

1v+1u=1f1v+1u=1f

12u+1u=1−3012u+1u=1−30

u=-45 cms. which is between the focal length and the Curvature.

For virtual image

m=−vu=2m=−vu=2

v=-2u

Now Using the mirror equation,

1v+1u=1f1v+1u=1f

1−2u+1u=1−301−2u+1u=1−30

u=-15 cm which is between the focal length and the pole

Answer 2

Answer:

$&\mathrm{u}=-15 \mathrm{~cm}$ be the distance of the object from the mirror.

Explanation:

Given:

Rohit placed a pencil perpendicular to a principal axis in front of a converging mirror (Another name Concave mirror) of a focal length of 30 cm. The image formed (h') is twice the size of the pencil (h). Calculate the distance of the object from the mirror (u).

f = - 30 cm

[We have given that mirror exists converging means concave mirror. And Image formed exists twice the size of the object. This means the image formed is behind the mirror and is larger in size. So, we take f in negative sign.]

h' = 2h

u = ?

Now;

$\mathrm{m}=\frac{-v}{u}=\frac{h^{\prime}}{h}$

$&\frac{-v}{u}=\frac{2 h}{h}$

$&\frac{-v}{u}=2$

$&-\mathrm{v}=2 \mathrm{u}$

$&\mathrm{v}=-2 \mathrm{u}$

As we know that

$&\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Substituting the values of f, u, and v in the above equation, we get

$&\frac{1}{-30}=\frac{1}{-2 u}+\frac{1}{u}$

$&\frac{-1}{30}=\frac{-1}{2 u}+\frac{-1+2}{2 u}$

$&\frac{-1}{30}=\frac{+1}{2 u}$

Simplifying the equation, we get

$&2 \mathrm{u}=-30 \mathrm{~cm}$

$&\mathrm{u}=-15 \mathrm{~cm}$

$&\mathrm{u}=-15 \mathrm{~cm}$, which is between the focal length and the pole.

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