Question

what is linear magnification

Answer 1

Hello mate

Here is your answer,

Linear magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

Answer 2

$\Huge\bf\purple{\mid{\overline{\underline{Answer}}}\mid}$

The ratio of the height of the image to the height of the object is known as linear magnification. If the value of ratio is more than unity, then image is magnified; if ratio is equal to unity, then image is of the same size as that of the object; and if ratio is less than unity, then image is diminished.

$\tt \green{Linear\:magnification(m)=\frac{Height\:of\:image(I)}{Height\:of\:object(O)} }$

Mathematically

$\large\tt \orange{m=\frac{I}{O} =-\frac{v}{u} }$

$\begin{lgathered}\bf{where}\begin{cases}\sf{v \dashrightarrow{}distance\:of\:image\:from\:the\:pole\:of\:the\:mirror.}\\ \sf{u \dashrightarrow{}distance\:of\:object\:from\:the\:pole\:of\:the\:mirror.}\end{cases}\end{lgathered}$

The ratio of object length to image length is 1:4. What will be the ratio of distance of object (u) to distance of image (v) from lens?

$\begin{lgathered}\bf{Given}\begin{cases}\sf{Object \dashrightarrow1}\\ \sf{Image \dashrightarrow4}\end{cases}\end{lgathered}$

$\large \tt{\frac{I}{O} =\frac{4}{1}}$

So by linear magnification(m) =$\bf{\frac{I}{O} =\frac{v}{u} }$

$\large \tt{\frac{v}{u} =\frac{4}{1}}$

Thus,

The required ratio is u:v = 1:4.

$\huge{\mathfrak{\orange{hope\; it\; helps}}}$