Question

an object 2 cm in size is placed at 3 cm in front of a concave mirror of focal length 15 centimetres at what distance from mirror should the screen be placed in order to obtain a sharp image find the nature and size of the image

Answer 1

$\bf {\huge {\bold{SOLUTION:-}}}$


$\bf{Given:-}$

Height of Object (ho)= 2 cm

Object distance (u) = - 3 cm

Focal Length (f) = -15

Here, Using Lens Formula!

Note: Lens Formula shows that Relationship between object, image Distance and focal length of the mirror.

Lens Formula:
$\bf{ \huge{ \fbox{ \mathsf { \frac{1}{f} = \frac{1}{v} + \frac{1}{u}}}}}$

Now, Substitute the Given value in Lens Formula So, We get!



$\bf{ \huge{ \fbox{ \mathsf { \frac{1}{ - 15} = \frac{1}{v} + \frac{1}{ - 3}}}}} \\ \\ \\ \bf{ \huge{ \fbox{ \mathsf { \frac{1}{ - 15} + \frac{1}{3} = \frac{1}{v}}}}} \\ \\ \mathsf{ \bf{ \fbox{ \huge{ \frac{1}{v} = \frac{ - 1 + 5}{ 15}}}}} \\ \\ \\ \mathsf{ \bf{ \fbox{ \huge{ \frac{1}{v} = \frac{ 4}{ 15}}}}} \\ \\ \mathsf{ \bf{ \fbox{ \huge{v = \frac{15}{4}}}}} \\ \\ \\ \mathsf{ \bf{ \fbox{ \fbox{ \huge{v = 3.75cm}}}}}$


Now, Using Magnification Formula!

$\mathsf{ \huge{ \bf{ \fbox{m = \frac{ {h}^{ `} }{ho} = \frac{ - v}{u}}}}}$

Now, Substitute the Given value in Magnification!


$\mathsf{ \huge{ \bf{ \fbox{ \frac{ {h}^{ } }{2} = \frac{ - 3.75}{ - 3 }}}}} \\ \\ \\ \\ \mathsf{ \huge{ \bf{ \fbox{h = \frac{7.5}{3} \implies2.5}}}}$


$\bold{\fbox{Hence \: \: Size \: \: of \: \: Image \: =2.5cm}}$



Conclusion:


The Image formed virtual and erect and behind the mirror at 2.5 cm.