When an object is placed at a distance of 25 cm from a mirror, the magnification is m1 . But the magnification becomes m2 , when the object is moved 15 cm farther away with respect to the earlier position. (m1/m2)=4 text, then find the focal length of the mirror and what type of mirror it is?

Question

When an object is placed at a distance of 25 cm from a mirror, the magnification is m1 . But the magnification becomes m2 , when the object is moved 15 cm farther away with respect to the earlier position. (m1/m2)=4 text, then find the focal length of the mirror and what type of mirror it is? (A) 20cm , convex (B) 20cm , concave (C) 10cm , convex (D) 10cm , concave

Tardigrade Admin · Accepted Answer

textitm=-( textitv/ textitu)=( textitf/ textitf - textitu) so, m1=(f/f-(-25))=(f/f+25) textSimilarly, textitm2=( textitf/ textitf + text40) textand( textitm1/ textitm2)=4 ∴ ( textitf + text40/ textitf + text25)=4 f=-20cm negative sign shows that the mirror is concave.

$20 c m$ , convex

$20 c m$ , concave

$10 c m$ , convex

$10 c m$ , concave

$m = - \frac{v}{u} = \frac{f}{f - u}$
so, $m_{1} = \frac{f}{f - ( - 25 )} = \frac{f}{f + 25}$
$Similarly, m_{2} = \frac{f}{f + 40}$
$and \frac{m _{1}}{m _{2}} = 4$
$∴ \frac{f + 40}{f + 25} = 4$
$f = - 20 c m$
negative sign shows that the mirror is concave.

20cm , convex

20cm , concave

10cm , convex

10cm , concave

m=−uv​=f−uf​ so, m1​=f−(−25)f​=f+25f​ Similarly,m2​=f+40f​ andm2​m1​​=4 ∴f+25f+40​=4 f=−20cm negative sign shows that the mirror is concave.

$20 c m$ , convex

$20 c m$ , concave

$10 c m$ , convex

$10 c m$ , concave

$m = - \frac{v}{u} = \frac{f}{f - u}$
so, $m_{1} = \frac{f}{f - ( - 25 )} = \frac{f}{f + 25}$
$Similarly, m_{2} = \frac{f}{f + 40}$
$and \frac{m _{1}}{m _{2}} = 4$
$∴ \frac{f + 40}{f + 25} = 4$
$f = - 20 c m$
negative sign shows that the mirror is concave.