Question

a 2.0 cm high object is placed 7.10 cm from a concave mirror whose radius of curvature is 10.20cm

Answer 1

Answer:

f=1/2R = ½(10.20)=5.10cm, so the object is located between the center of curvature and the focal point.

Answer 2

Answer:

The location of the image is$18.105 \mathrm{~cm}$ and the size of the image is

$5.1 \mathrm{~cm}$.

Explanation:

Step 1: Given data and assumptions

Height of object$=h_0=2 \mathrm{~cm}$

Distance between object and mirror $=u=7.1 \mathrm{~cm}$

Radius of curvature $=R=10.2 \mathrm{~cm}$

Focal length $=\frac{R}{2}=\frac{10.2}{2}=5.1 \mathrm{~cm}$

Let the location of the image be v and the size of the image be h

For the concave mirror, the sign convention is

$\begin{aligned}& f=-5.1 \mathrm{~cm}[-\text { ve sign }] \\& u=-7.1 \mathrm{~cm}[-\text { ve sign }] \\& h_0=2 \mathrm{~cm}[+\text { ve sign }]\end{aligned}$

Step 2: Finding the value of v by using the mirror formula

Using the mirror formula,

$\begin{aligned}& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\& \Rightarrow \frac{1}{-5.1}=\frac{1}{v}-\frac{1}{7.1} \\& \Rightarrow \frac{1}{7.1}-\frac{1}{5.1}=\frac{1}{v} \\& \Rightarrow \frac{1}{v}=-\frac{2}{36.21} \\& \Rightarrow v=-18.105\end{aligned}$

Step 3 : Finding the required value by using the magnification formula

Using magnification formula

$\begin{aligned}m & =-\frac{v}{u}=\frac{h}{h_0} \\& \Rightarrow-\frac{-18.105}{-7.1}=\frac{h}{2} \\& \Rightarrow-2.55=\frac{h}{2} \\& \Rightarrow h=-5.1 \mathrm{~cm}\end{aligned}$

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