An object is placed at a distance of 10, cm f | vedclass

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Question

(A) Focal length of convex mirror,$f = +15\, cm$.Object distance,$u = -10\, cm$.According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{

Vedclass · Accepted Answer

$6\, cm$ behind the mirror,virtual and erect

Vedclass · Answer

(A) Focal length of convex mirror,$f = +15\, cm$.Object distance,$u = -10\, cm$.According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Substituting the values,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-10}) = \frac{1}{15} + \frac{1}{10}$.$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$.Therefore,$v = +6\, cm$.The positive value of $v$ indicates that the image is formed $6\, cm$ behind the mirror.Magnification,$m = -\frac{v}{u} = -(\frac{6}{-10}) = +0.6$.The positive value of magnification indicates that the image formed is virtual and erect.

Material medium	Refractive index
Water	$1.33$
Kerosene	$1.44$
Turpentine oil	$1.47$

An object is placed at a distance of $10\, cm$ from a convex mirror of focal length $15\, cm$. Find the position and nature of the image.

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