An object is placed at a distance of 50 , cm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The object is at a distance of $50 \, cm$ from the convex mirror. The plane mirror is placed at a distance of $30 \, cm$ from the object.Since the

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) The object is at a distance of $50 \, cm$ from the convex mirror. The plane mirror is placed at a distance of $30 \, cm$ from the object.Since the object is $30 \, cm$ in front of the plane mirror,the image formed by the plane mirror is $30 \, cm$ behind it.From the diagram,the distance of the plane mirror from the pole $P$ of the convex mirror is $50 \, cm - 30 \, cm = 20 \, cm$.Therefore,the image formed by the plane mirror is at a distance of $30 \, cm - 20 \, cm = 10 \, cm$ behind the convex mirror (i.e.,$v = +10 \, cm$).Since there is no parallax,the image formed by the convex mirror must coincide with the image formed by the plane mirror.For the convex mirror,the object distance $u = -50 \, cm$ and the image distance $v = +10 \, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{10} + \frac{1}{-50} = \frac{1}{f}$$\frac{5 - 1}{50} = \frac{1}{f} \Rightarrow \frac{4}{50} = \frac{1}{f} \Rightarrow f = 12.5 \, cm$.The radius of curvature $R = 2f = 2 	imes 12.5 = 25 \, cm$.

Similar Questions

Nuclear fusion is common to the pair:

$A$ wire has a resistance of $12 \, \Omega$. It is bent in the form of an equilateral triangle. The effective resistance between any two corners of the triangle is:

Two of the lines represented by the equation $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$ will be perpendicular,then

Alzheimer's disease in humans is associated with the deficiency of:

If $a \cos^3 \alpha + 3a \cos \alpha \sin^2 \alpha = m$ and $a \sin^3 \alpha + 3a \cos^2 \alpha \sin \alpha = n$,then $(m + n)^{2/3} + (m - n)^{2/3}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module