Question

The real image 4 times magnified than the object is produced by a concave mirror of focal lenght 30cm. Calculate the distance of the object from the mirror.
Class 10
Physics
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Answer 1

Answer:

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Answer 2

Solution :

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

• Magnification, (m) = 4
• Focal length of the mirror, (f) = -30 cm

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

As we know that formula of the magnification;

$\mapsto\bf{m=\dfrac{Height\:of\:Image\:(I)}{Height\:of\:object\:(0)} =\dfrac{Distance\:of\:Image}{Distance\:of\:Object} =\dfrac{v}{u} }$

A/q

When Image is real & inverted, magnification will be negative .

$\mapsto\tt{m=\dfrac{-v}{u} }$

$\mapsto\tt{-4=\dfrac{-v}{u} }$

$\mapsto\tt{\cancel{-}v=\cancel{-}4u}$

$\mapsto\tt{v=4u}$

Now, using formula of the mirror;

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

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

$\mapsto\tt{\dfrac{1}{-30} =\dfrac{1}{4u}+\dfrac{1}{u}}$

$\mapsto\tt{\dfrac{1}{-30} =\dfrac{1+4}{4u}}$

$\mapsto\tt{\dfrac{1}{-30} =\dfrac{5}{4u}}$

$\mapsto\tt{4u = -30\times 5 \:\:\underbrace{\sf{cross-multiplication}}}$

$\mapsto\tt{4u=-150}$

$\mapsto\tt{u=-\cancel{150/4}}$

$\mapsto\bf{u=-37.5\:cm}$

Thus;

The distance of the object from the the mirror will be 37.5 cm in front of the mirror .