282013
The light rays from an object have bee reflected towards an observer from a standar flat mirror, the image observed by the observe are:-
1 Real
2 Eract
3 Smaller in size then object
4 Laterally inverted
Explanation:
C: Properties of an image formed by plane mirror.
The image obtained is virtual
The image is laterally inverted
The image is erect
The size of the image is the same as the size of the object the distance between the image obtained is the same as the distance between the object from the mirror.
JEE Main-29.01.2023
Ray Optics
282014
The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of an material of R.I. 1.6, the radii 0 curvature of the lens is
1 $-2.66 \mathrm{~cm}$
2 $115.44 \mathrm{~cm}$
3 $-26.6 \mathrm{~cm}$
4 $+36.6 \mathrm{~cm}$
Explanation:
C: Given, Refractive index $(\mu)=1.6$
Power of lens $(P)=-4.5$
Let, $R$ is the radius of curvature of equal concave lens, Then, $\mathrm{R}_1=-\mathrm{R}, \mathrm{R}_2=+\mathrm{R}$
We know that,
or
$\begin{aligned}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{-1}{4.5}=\frac{-2}{9} \mathrm{~m}
\end{aligned}$
And
$\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=(1.6-1)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{-9}{2}=\frac{-0.6 \times 2}{\mathrm{R}} \\
\mathrm{R}=\frac{0.6 \times 2 \times 2}{9} \\
\mathrm{R}=0.266 \mathrm{~m} \\
\mathrm{R}=26.7 \mathrm{~cm}
\end{aligned}$
Karnataka CET-2022
Ray Optics
282015
The power of a corrective lens is $-4.0 \mathrm{D}$. The lens is
1 convex lens of focal length $+25 \mathrm{~cm}$
2 concave lens of focal length $-25 \mathrm{~cm}$
3 convex lens of focal length $+4 \mathrm{~cm}$
4 concave lens of focal length $-4 \mathrm{~cm}$
(e) convex lens of focal length $+20 \mathrm{~cm}$
Explanation:
B: Given, power of lens (P) $=-4 \mathrm{D}$
We know that,
Power $(\mathrm{P})=\frac{1}{\mathrm{f}}$
Where, $f=$ focal length
$\begin{aligned}
\mathrm{f}=-\frac{1}{4}=-0.25 \mathrm{~m} \\
\mathrm{f}=-25 \mathrm{~cm}
\end{aligned}$
Kerala CEE 04.07.2022
Ray Optics
282016
The image formed by a convex mirror is only $\frac{1}{3}$ of the size of the object. If the focal length of the mirror is $30 \mathrm{~cm}$, the image will be formed with reference to the mirror at
1 $12 \mathrm{~cm}$ behind
2 $10 \mathrm{~cm}$ in front
3 $20 \mathrm{~cm}$ behind
4 $20 \mathrm{~cm}$ in front
Explanation:
C: Given, focal length (f) $=30 \mathrm{~cm}$
And, height of image $\left(h_2\right)=\frac{\text { height of object }\left(h_1\right)}{3}$
$\mathrm{h}_2=\frac{\mathrm{h}_1}{3}$
And magnification $(\mathrm{m})=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}}$
$\begin{aligned}
\frac{\frac{\mathrm{h}_1}{3}}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}} \\
\mathrm{u}=-3 \mathrm{v}
\end{aligned}$
From mirror formula -
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\begin{aligned}
\frac{1}{30}=\frac{1}{-3 \mathrm{v}}+\frac{1}{\mathrm{v}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1}{3}+1\right) \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1+3}{3}\right) \\
\frac{1}{30}=\frac{2}{3 \mathrm{v}} \\
\mathrm{v}=20 \mathrm{~cm}
\end{aligned}$
Hence, image is formed $20 \mathrm{~cm}$ behind the mirror.
Assam CEE-31.07.2022
Ray Optics
282017
An object is placed at $10 \mathrm{~cm}$ in front of a concave mirror. If the image is at $20 \mathrm{~cm}$ from the mirror on the same side of the object, then magnification produced by the mirror is
282013
The light rays from an object have bee reflected towards an observer from a standar flat mirror, the image observed by the observe are:-
1 Real
2 Eract
3 Smaller in size then object
4 Laterally inverted
Explanation:
C: Properties of an image formed by plane mirror.
The image obtained is virtual
The image is laterally inverted
The image is erect
The size of the image is the same as the size of the object the distance between the image obtained is the same as the distance between the object from the mirror.
JEE Main-29.01.2023
Ray Optics
282014
The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of an material of R.I. 1.6, the radii 0 curvature of the lens is
1 $-2.66 \mathrm{~cm}$
2 $115.44 \mathrm{~cm}$
3 $-26.6 \mathrm{~cm}$
4 $+36.6 \mathrm{~cm}$
Explanation:
C: Given, Refractive index $(\mu)=1.6$
Power of lens $(P)=-4.5$
Let, $R$ is the radius of curvature of equal concave lens, Then, $\mathrm{R}_1=-\mathrm{R}, \mathrm{R}_2=+\mathrm{R}$
We know that,
or
$\begin{aligned}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{-1}{4.5}=\frac{-2}{9} \mathrm{~m}
\end{aligned}$
And
$\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=(1.6-1)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{-9}{2}=\frac{-0.6 \times 2}{\mathrm{R}} \\
\mathrm{R}=\frac{0.6 \times 2 \times 2}{9} \\
\mathrm{R}=0.266 \mathrm{~m} \\
\mathrm{R}=26.7 \mathrm{~cm}
\end{aligned}$
Karnataka CET-2022
Ray Optics
282015
The power of a corrective lens is $-4.0 \mathrm{D}$. The lens is
1 convex lens of focal length $+25 \mathrm{~cm}$
2 concave lens of focal length $-25 \mathrm{~cm}$
3 convex lens of focal length $+4 \mathrm{~cm}$
4 concave lens of focal length $-4 \mathrm{~cm}$
(e) convex lens of focal length $+20 \mathrm{~cm}$
Explanation:
B: Given, power of lens (P) $=-4 \mathrm{D}$
We know that,
Power $(\mathrm{P})=\frac{1}{\mathrm{f}}$
Where, $f=$ focal length
$\begin{aligned}
\mathrm{f}=-\frac{1}{4}=-0.25 \mathrm{~m} \\
\mathrm{f}=-25 \mathrm{~cm}
\end{aligned}$
Kerala CEE 04.07.2022
Ray Optics
282016
The image formed by a convex mirror is only $\frac{1}{3}$ of the size of the object. If the focal length of the mirror is $30 \mathrm{~cm}$, the image will be formed with reference to the mirror at
1 $12 \mathrm{~cm}$ behind
2 $10 \mathrm{~cm}$ in front
3 $20 \mathrm{~cm}$ behind
4 $20 \mathrm{~cm}$ in front
Explanation:
C: Given, focal length (f) $=30 \mathrm{~cm}$
And, height of image $\left(h_2\right)=\frac{\text { height of object }\left(h_1\right)}{3}$
$\mathrm{h}_2=\frac{\mathrm{h}_1}{3}$
And magnification $(\mathrm{m})=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}}$
$\begin{aligned}
\frac{\frac{\mathrm{h}_1}{3}}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}} \\
\mathrm{u}=-3 \mathrm{v}
\end{aligned}$
From mirror formula -
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\begin{aligned}
\frac{1}{30}=\frac{1}{-3 \mathrm{v}}+\frac{1}{\mathrm{v}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1}{3}+1\right) \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1+3}{3}\right) \\
\frac{1}{30}=\frac{2}{3 \mathrm{v}} \\
\mathrm{v}=20 \mathrm{~cm}
\end{aligned}$
Hence, image is formed $20 \mathrm{~cm}$ behind the mirror.
Assam CEE-31.07.2022
Ray Optics
282017
An object is placed at $10 \mathrm{~cm}$ in front of a concave mirror. If the image is at $20 \mathrm{~cm}$ from the mirror on the same side of the object, then magnification produced by the mirror is
282013
The light rays from an object have bee reflected towards an observer from a standar flat mirror, the image observed by the observe are:-
1 Real
2 Eract
3 Smaller in size then object
4 Laterally inverted
Explanation:
C: Properties of an image formed by plane mirror.
The image obtained is virtual
The image is laterally inverted
The image is erect
The size of the image is the same as the size of the object the distance between the image obtained is the same as the distance between the object from the mirror.
JEE Main-29.01.2023
Ray Optics
282014
The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of an material of R.I. 1.6, the radii 0 curvature of the lens is
1 $-2.66 \mathrm{~cm}$
2 $115.44 \mathrm{~cm}$
3 $-26.6 \mathrm{~cm}$
4 $+36.6 \mathrm{~cm}$
Explanation:
C: Given, Refractive index $(\mu)=1.6$
Power of lens $(P)=-4.5$
Let, $R$ is the radius of curvature of equal concave lens, Then, $\mathrm{R}_1=-\mathrm{R}, \mathrm{R}_2=+\mathrm{R}$
We know that,
or
$\begin{aligned}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{-1}{4.5}=\frac{-2}{9} \mathrm{~m}
\end{aligned}$
And
$\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=(1.6-1)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{-9}{2}=\frac{-0.6 \times 2}{\mathrm{R}} \\
\mathrm{R}=\frac{0.6 \times 2 \times 2}{9} \\
\mathrm{R}=0.266 \mathrm{~m} \\
\mathrm{R}=26.7 \mathrm{~cm}
\end{aligned}$
Karnataka CET-2022
Ray Optics
282015
The power of a corrective lens is $-4.0 \mathrm{D}$. The lens is
1 convex lens of focal length $+25 \mathrm{~cm}$
2 concave lens of focal length $-25 \mathrm{~cm}$
3 convex lens of focal length $+4 \mathrm{~cm}$
4 concave lens of focal length $-4 \mathrm{~cm}$
(e) convex lens of focal length $+20 \mathrm{~cm}$
Explanation:
B: Given, power of lens (P) $=-4 \mathrm{D}$
We know that,
Power $(\mathrm{P})=\frac{1}{\mathrm{f}}$
Where, $f=$ focal length
$\begin{aligned}
\mathrm{f}=-\frac{1}{4}=-0.25 \mathrm{~m} \\
\mathrm{f}=-25 \mathrm{~cm}
\end{aligned}$
Kerala CEE 04.07.2022
Ray Optics
282016
The image formed by a convex mirror is only $\frac{1}{3}$ of the size of the object. If the focal length of the mirror is $30 \mathrm{~cm}$, the image will be formed with reference to the mirror at
1 $12 \mathrm{~cm}$ behind
2 $10 \mathrm{~cm}$ in front
3 $20 \mathrm{~cm}$ behind
4 $20 \mathrm{~cm}$ in front
Explanation:
C: Given, focal length (f) $=30 \mathrm{~cm}$
And, height of image $\left(h_2\right)=\frac{\text { height of object }\left(h_1\right)}{3}$
$\mathrm{h}_2=\frac{\mathrm{h}_1}{3}$
And magnification $(\mathrm{m})=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}}$
$\begin{aligned}
\frac{\frac{\mathrm{h}_1}{3}}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}} \\
\mathrm{u}=-3 \mathrm{v}
\end{aligned}$
From mirror formula -
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\begin{aligned}
\frac{1}{30}=\frac{1}{-3 \mathrm{v}}+\frac{1}{\mathrm{v}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1}{3}+1\right) \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1+3}{3}\right) \\
\frac{1}{30}=\frac{2}{3 \mathrm{v}} \\
\mathrm{v}=20 \mathrm{~cm}
\end{aligned}$
Hence, image is formed $20 \mathrm{~cm}$ behind the mirror.
Assam CEE-31.07.2022
Ray Optics
282017
An object is placed at $10 \mathrm{~cm}$ in front of a concave mirror. If the image is at $20 \mathrm{~cm}$ from the mirror on the same side of the object, then magnification produced by the mirror is
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Ray Optics
282013
The light rays from an object have bee reflected towards an observer from a standar flat mirror, the image observed by the observe are:-
1 Real
2 Eract
3 Smaller in size then object
4 Laterally inverted
Explanation:
C: Properties of an image formed by plane mirror.
The image obtained is virtual
The image is laterally inverted
The image is erect
The size of the image is the same as the size of the object the distance between the image obtained is the same as the distance between the object from the mirror.
JEE Main-29.01.2023
Ray Optics
282014
The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of an material of R.I. 1.6, the radii 0 curvature of the lens is
1 $-2.66 \mathrm{~cm}$
2 $115.44 \mathrm{~cm}$
3 $-26.6 \mathrm{~cm}$
4 $+36.6 \mathrm{~cm}$
Explanation:
C: Given, Refractive index $(\mu)=1.6$
Power of lens $(P)=-4.5$
Let, $R$ is the radius of curvature of equal concave lens, Then, $\mathrm{R}_1=-\mathrm{R}, \mathrm{R}_2=+\mathrm{R}$
We know that,
or
$\begin{aligned}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{-1}{4.5}=\frac{-2}{9} \mathrm{~m}
\end{aligned}$
And
$\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=(1.6-1)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{-9}{2}=\frac{-0.6 \times 2}{\mathrm{R}} \\
\mathrm{R}=\frac{0.6 \times 2 \times 2}{9} \\
\mathrm{R}=0.266 \mathrm{~m} \\
\mathrm{R}=26.7 \mathrm{~cm}
\end{aligned}$
Karnataka CET-2022
Ray Optics
282015
The power of a corrective lens is $-4.0 \mathrm{D}$. The lens is
1 convex lens of focal length $+25 \mathrm{~cm}$
2 concave lens of focal length $-25 \mathrm{~cm}$
3 convex lens of focal length $+4 \mathrm{~cm}$
4 concave lens of focal length $-4 \mathrm{~cm}$
(e) convex lens of focal length $+20 \mathrm{~cm}$
Explanation:
B: Given, power of lens (P) $=-4 \mathrm{D}$
We know that,
Power $(\mathrm{P})=\frac{1}{\mathrm{f}}$
Where, $f=$ focal length
$\begin{aligned}
\mathrm{f}=-\frac{1}{4}=-0.25 \mathrm{~m} \\
\mathrm{f}=-25 \mathrm{~cm}
\end{aligned}$
Kerala CEE 04.07.2022
Ray Optics
282016
The image formed by a convex mirror is only $\frac{1}{3}$ of the size of the object. If the focal length of the mirror is $30 \mathrm{~cm}$, the image will be formed with reference to the mirror at
1 $12 \mathrm{~cm}$ behind
2 $10 \mathrm{~cm}$ in front
3 $20 \mathrm{~cm}$ behind
4 $20 \mathrm{~cm}$ in front
Explanation:
C: Given, focal length (f) $=30 \mathrm{~cm}$
And, height of image $\left(h_2\right)=\frac{\text { height of object }\left(h_1\right)}{3}$
$\mathrm{h}_2=\frac{\mathrm{h}_1}{3}$
And magnification $(\mathrm{m})=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}}$
$\begin{aligned}
\frac{\frac{\mathrm{h}_1}{3}}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}} \\
\mathrm{u}=-3 \mathrm{v}
\end{aligned}$
From mirror formula -
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\begin{aligned}
\frac{1}{30}=\frac{1}{-3 \mathrm{v}}+\frac{1}{\mathrm{v}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1}{3}+1\right) \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1+3}{3}\right) \\
\frac{1}{30}=\frac{2}{3 \mathrm{v}} \\
\mathrm{v}=20 \mathrm{~cm}
\end{aligned}$
Hence, image is formed $20 \mathrm{~cm}$ behind the mirror.
Assam CEE-31.07.2022
Ray Optics
282017
An object is placed at $10 \mathrm{~cm}$ in front of a concave mirror. If the image is at $20 \mathrm{~cm}$ from the mirror on the same side of the object, then magnification produced by the mirror is
282013
The light rays from an object have bee reflected towards an observer from a standar flat mirror, the image observed by the observe are:-
1 Real
2 Eract
3 Smaller in size then object
4 Laterally inverted
Explanation:
C: Properties of an image formed by plane mirror.
The image obtained is virtual
The image is laterally inverted
The image is erect
The size of the image is the same as the size of the object the distance between the image obtained is the same as the distance between the object from the mirror.
JEE Main-29.01.2023
Ray Optics
282014
The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of an material of R.I. 1.6, the radii 0 curvature of the lens is
1 $-2.66 \mathrm{~cm}$
2 $115.44 \mathrm{~cm}$
3 $-26.6 \mathrm{~cm}$
4 $+36.6 \mathrm{~cm}$
Explanation:
C: Given, Refractive index $(\mu)=1.6$
Power of lens $(P)=-4.5$
Let, $R$ is the radius of curvature of equal concave lens, Then, $\mathrm{R}_1=-\mathrm{R}, \mathrm{R}_2=+\mathrm{R}$
We know that,
or
$\begin{aligned}
\mathrm{P}=\frac{1}{\mathrm{f}} \\
\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{-1}{4.5}=\frac{-2}{9} \mathrm{~m}
\end{aligned}$
And
$\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right) \\
\frac{1}{\mathrm{f}}=(1.6-1)\left(-\frac{2}{\mathrm{R}}\right) \\
\frac{-9}{2}=\frac{-0.6 \times 2}{\mathrm{R}} \\
\mathrm{R}=\frac{0.6 \times 2 \times 2}{9} \\
\mathrm{R}=0.266 \mathrm{~m} \\
\mathrm{R}=26.7 \mathrm{~cm}
\end{aligned}$
Karnataka CET-2022
Ray Optics
282015
The power of a corrective lens is $-4.0 \mathrm{D}$. The lens is
1 convex lens of focal length $+25 \mathrm{~cm}$
2 concave lens of focal length $-25 \mathrm{~cm}$
3 convex lens of focal length $+4 \mathrm{~cm}$
4 concave lens of focal length $-4 \mathrm{~cm}$
(e) convex lens of focal length $+20 \mathrm{~cm}$
Explanation:
B: Given, power of lens (P) $=-4 \mathrm{D}$
We know that,
Power $(\mathrm{P})=\frac{1}{\mathrm{f}}$
Where, $f=$ focal length
$\begin{aligned}
\mathrm{f}=-\frac{1}{4}=-0.25 \mathrm{~m} \\
\mathrm{f}=-25 \mathrm{~cm}
\end{aligned}$
Kerala CEE 04.07.2022
Ray Optics
282016
The image formed by a convex mirror is only $\frac{1}{3}$ of the size of the object. If the focal length of the mirror is $30 \mathrm{~cm}$, the image will be formed with reference to the mirror at
1 $12 \mathrm{~cm}$ behind
2 $10 \mathrm{~cm}$ in front
3 $20 \mathrm{~cm}$ behind
4 $20 \mathrm{~cm}$ in front
Explanation:
C: Given, focal length (f) $=30 \mathrm{~cm}$
And, height of image $\left(h_2\right)=\frac{\text { height of object }\left(h_1\right)}{3}$
$\mathrm{h}_2=\frac{\mathrm{h}_1}{3}$
And magnification $(\mathrm{m})=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}}$
$\begin{aligned}
\frac{\frac{\mathrm{h}_1}{3}}{\mathrm{~h}_1}=\frac{-\mathrm{v}}{\mathrm{u}} \\
\mathrm{u}=-3 \mathrm{v}
\end{aligned}$
From mirror formula -
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\begin{aligned}
\frac{1}{30}=\frac{1}{-3 \mathrm{v}}+\frac{1}{\mathrm{v}} \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1}{3}+1\right) \\
\frac{1}{30}=\frac{1}{\mathrm{v}}\left(\frac{-1+3}{3}\right) \\
\frac{1}{30}=\frac{2}{3 \mathrm{v}} \\
\mathrm{v}=20 \mathrm{~cm}
\end{aligned}$
Hence, image is formed $20 \mathrm{~cm}$ behind the mirror.
Assam CEE-31.07.2022
Ray Optics
282017
An object is placed at $10 \mathrm{~cm}$ in front of a concave mirror. If the image is at $20 \mathrm{~cm}$ from the mirror on the same side of the object, then magnification produced by the mirror is