Question

focal length of concave mirror is 30 cm if an object is placed at 40cm give the position of image determine the magnification of the image​

Answer 1

Answer:

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Explanation:

We have f= -30 and u= -40 then position of the image... 1/f = 1/ v+1/ u . We put the value of u& f we get -1/30=1/v-1/40 this is equal to 1/v= -1/12

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Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Focal length from mirror, (f) = -30 cm
• Distance of object from mirror, (u) = -40 cm

$\underline{\bf{Explanation\::}}$

As we know that formula of the mirror;

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

A/q

$\mapsto\tt{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }$

$\mapsto\tt{\dfrac{1}{(-30)} = \dfrac{1}{v} + \dfrac{1}{(-40)} }$

$\mapsto\tt{\dfrac{1}{(-30)} = \dfrac{1}{v} -\dfrac{1}{40} }$

$\mapsto\tt{\dfrac{1}{v} = -\dfrac{1}{30} +\dfrac{1}{40} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-40 + 30}{120} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-10}{120} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-1\cancel{0}}{12\cancel{0}} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-1}{12} }$

$\mapsto\bf{v = -12\:cm}$

Thus,

The position of image or distance of image will be 12 cm .

As we know that formula of the magnification;

$\bf{m=\dfrac{Height\:of\:Image \:(I)}{Height\:of\:object\:(O)} = \dfrac{Distance\:of\:Image }{Distance\:of\:object}=\dfrac{v}{u} }$

Now,

$\mapsto\tt{Magnification\:,(m) = \dfrac{v}{u} }$

$\mapsto\tt{Magnification\:,(m) = \dfrac{-12}{-40} }$

$\mapsto\tt{Magnification\:,(m) = \cancel{\dfrac{-12}{-40} }}$

$\mapsto\bf{Magnification\:,(m) = 0.3\:cm}$

Thus,

The magnification of the image will be 0.3 cm .