Question

In order to produce an image with magnification m, a concave mirror of focal length f is used. If the object is kept at a distance d from the mirror, then $(m–1)f$ equals

• A. $md$
• B. $\frac{m}{(d+1)}$
• C. $\frac{m}{d}$
• D. $\frac{1}{md}$

Answer 1

The correct option is A

$md$

STEP 1: Given data

In a concave mirror

The object's distance from the mirror pole i.e. u $=-d$ (since object is placed on the left side of mirror, hence distance is taken as negative)

The focal length of the concave mirror is always negative$=-f$.

(The distance calculated in the direction of the incident ray is taken as positive and the opposite direction is taken as negative.)

STEP 2: Finding the position of the image

Using the mirror formula,

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

$\frac{1}{v}=\frac{1}{-f}-\frac{1}{-u}$

$\frac{1}{v}=-\frac{1}{f}+\frac{1}{d}$

$\frac{1}{v}=\left(\frac{f-d}{fd}\right)$

The image distance from the mirror i.e. $v=\left(\frac{df}{f-d}\right)$.

STEP 3: Finding the magnification

Using the magnification formula in the mirror

$m=-\frac{v}{u}$

$m=-\frac{v}{(-d)}$

$m=\left(\frac{fd}{(f-d)d}\right)$

$m=\left(\frac{f}{f-d}\right)$

STEP 4: Putting the values in the expression given

$\left(m-1\right)f=\left(\left(\frac{f}{f-d}\right)-1\right)f$

$\left(m-1\right)f=\left(\frac{f-f+d}{f-d}\right)f$

$\left(m-1\right)f=\left(\frac{df}{f-d}\right)$

Replacing the expression on RHS with v;

$\left(m-1\right)f=-v$

Since $m=-\frac{v}{d}$

$md=-v$

$(m-1)f=md$

Therefore, the correct option is (A).