Question

an object 4cm in size is placed at 25cm infront of a convex mirror of focal length 15cm at what distance from mirror should a screen be placed in order to obtain a sharp image find the nature and the size of the image​

Answer 1

Answer:

Explanation:

Screen should be placed at dist. 37.5 cm from mirror......nature of image real and inverted...and magnification is -1.5....and height of object is -6 cm...please give me like if it’s correct....

Answer 2

According to sign convention:

focal length (f) = -15cm

object distance (u) = -25cm

object height (h) = +4cm

image distance (v)= ?

image height (h.) = ?

Substitute the above values in the equation

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

$\frac{1}{ - 15} = \frac{1}{v} + \frac{1}{ - 25} \\ \\ \frac{1}{v} = \frac{1}{25} - \frac{1}{15} \\ \\ \frac{1}{v} = \frac{ - 2}{75} \\ \\ v = - 37.5cm$

So the screen should be placed at 37.5 cm from the pole of the mirror. The image is real.

$Magnification \: ({m}) = \frac{hi}{ho} = \frac{ - v}{u}$

By substituting the above values

$\frac{hi}{4} = \frac{ - 37.5}{ - 25} \\ \\ hi = \frac{37.5}{25} \\ \\ hi = - 6cm$

So, the image is inverted and enlarged.