Text Solution
Solution : Here, `u = ?, f = 18 cm, v = +- 36 cm`, <br> `m = ?` <br> From `(1)/(v) - (1)/(u) = (1)/(f)`, `-(1)/(u) = (1)/(f) -(1)/(u) = (1)/(18) - (1)/(+- 36) = (2-(+-1))/(36)` <br> `:. u =-36 cm` and `u = -12 cm`. <br> `{:("When u"= -36 cm),(and v = 36 cm):}],m = (v)/(u) = (+36)/(-36) = -1`, Image is real and inverted. <br> `{:("When u" = -12 cm),( and v = -36 cm):}], m = (v)/(u) = (-36)/(-12) = +3`, Image is virtual and erect.
Given,
The focal length of a convex lens, f = 18 cm.
Image distance, v = 24 cm
Object distance, u = ?
To find- Magnification
Solution:
By using lens formula-
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$
where, v = image distance, u = object distance, and f = focal length
Substituting the values of f, v and u we get,
$\frac{1}{24}-\frac{1}{u}=\frac{1}{18}\phantom{\rule{0ex}{0ex}}$
$\frac{1}{24}-\frac{1}{18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$
$\frac{18-24}{24\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$
$\frac{6}{24\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$
$\frac{1}{4\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$
$u=-72cm\phantom{\rule{0ex}{0ex}}$
So, the object distance is -72cm.
The object should be placed at a distance of -72 cm from the lens.
Now, the equation for finding magnification of a lens can be given as-
$m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}$
Substituting the values in magnification formula we get-
$m=\frac{24}{-72}\phantom{\rule{0ex}{0ex}}$
$m=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
Hence, the magnification produced will be $m=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}$