364714
A linear object of height \(10\;cm\) is kept in front of a concave mirror of radius of curvature \(15\;cm\), at a distance of \(10\;cm\). The image formed is
364717
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15\,cm,\) the focal length of the mirror must be
1 \( - 10\,cm\)
2 \( - 12\,cm\)
3 \(10/3\,cm\)
4 \(15\,cm\)
Explanation:
Given: \(u+v=15\) \(v=15-u ; m=2=-\dfrac{v}{u}\) and \(u=-u\)
\(2=\dfrac{-(15-u)}{-u}\) \(2 u=+15-u\) \(\Rightarrow 3 u=15\) \( \Rightarrow u = 5\;cm\,;\,\,\,v = 15 - 5 = 10\;cm\) By mirror formula \(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-5}=\dfrac{1-2}{10}\) \( \Rightarrow f = - 10\;cm\)
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364718
An object \(2.5\;cm\) high placed at a distance of \(10\;cm\) from a concave mirror of radius of curvature \(30\;cm\) . The size of the image is
1 \(10.5\;cm\)
2 \(9.2\;cm\)
3 \(7.5\;cm\)
4 \(5.6\;cm\)
Explanation:
\(R = - 30cm \Rightarrow f = - 15cm,\) \(h = + 2.5cm,\,u = - 10cm.\) By mirror formula \(\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{\left( { - 10} \right)}} \Rightarrow v = 30cm\) Also \(\frac{{h'}}{h} = - \frac{v}{u} \Rightarrow \frac{1}{{\left( { + 2.5} \right)}} = \frac{{30}}{{\left( { - 10} \right)}} \Rightarrow I = + 7.5cm\)
364714
A linear object of height \(10\;cm\) is kept in front of a concave mirror of radius of curvature \(15\;cm\), at a distance of \(10\;cm\). The image formed is
364717
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15\,cm,\) the focal length of the mirror must be
1 \( - 10\,cm\)
2 \( - 12\,cm\)
3 \(10/3\,cm\)
4 \(15\,cm\)
Explanation:
Given: \(u+v=15\) \(v=15-u ; m=2=-\dfrac{v}{u}\) and \(u=-u\)
\(2=\dfrac{-(15-u)}{-u}\) \(2 u=+15-u\) \(\Rightarrow 3 u=15\) \( \Rightarrow u = 5\;cm\,;\,\,\,v = 15 - 5 = 10\;cm\) By mirror formula \(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-5}=\dfrac{1-2}{10}\) \( \Rightarrow f = - 10\;cm\)
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364718
An object \(2.5\;cm\) high placed at a distance of \(10\;cm\) from a concave mirror of radius of curvature \(30\;cm\) . The size of the image is
1 \(10.5\;cm\)
2 \(9.2\;cm\)
3 \(7.5\;cm\)
4 \(5.6\;cm\)
Explanation:
\(R = - 30cm \Rightarrow f = - 15cm,\) \(h = + 2.5cm,\,u = - 10cm.\) By mirror formula \(\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{\left( { - 10} \right)}} \Rightarrow v = 30cm\) Also \(\frac{{h'}}{h} = - \frac{v}{u} \Rightarrow \frac{1}{{\left( { + 2.5} \right)}} = \frac{{30}}{{\left( { - 10} \right)}} \Rightarrow I = + 7.5cm\)
364714
A linear object of height \(10\;cm\) is kept in front of a concave mirror of radius of curvature \(15\;cm\), at a distance of \(10\;cm\). The image formed is
364717
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15\,cm,\) the focal length of the mirror must be
1 \( - 10\,cm\)
2 \( - 12\,cm\)
3 \(10/3\,cm\)
4 \(15\,cm\)
Explanation:
Given: \(u+v=15\) \(v=15-u ; m=2=-\dfrac{v}{u}\) and \(u=-u\)
\(2=\dfrac{-(15-u)}{-u}\) \(2 u=+15-u\) \(\Rightarrow 3 u=15\) \( \Rightarrow u = 5\;cm\,;\,\,\,v = 15 - 5 = 10\;cm\) By mirror formula \(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-5}=\dfrac{1-2}{10}\) \( \Rightarrow f = - 10\;cm\)
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364718
An object \(2.5\;cm\) high placed at a distance of \(10\;cm\) from a concave mirror of radius of curvature \(30\;cm\) . The size of the image is
1 \(10.5\;cm\)
2 \(9.2\;cm\)
3 \(7.5\;cm\)
4 \(5.6\;cm\)
Explanation:
\(R = - 30cm \Rightarrow f = - 15cm,\) \(h = + 2.5cm,\,u = - 10cm.\) By mirror formula \(\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{\left( { - 10} \right)}} \Rightarrow v = 30cm\) Also \(\frac{{h'}}{h} = - \frac{v}{u} \Rightarrow \frac{1}{{\left( { + 2.5} \right)}} = \frac{{30}}{{\left( { - 10} \right)}} \Rightarrow I = + 7.5cm\)
364714
A linear object of height \(10\;cm\) is kept in front of a concave mirror of radius of curvature \(15\;cm\), at a distance of \(10\;cm\). The image formed is
364717
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15\,cm,\) the focal length of the mirror must be
1 \( - 10\,cm\)
2 \( - 12\,cm\)
3 \(10/3\,cm\)
4 \(15\,cm\)
Explanation:
Given: \(u+v=15\) \(v=15-u ; m=2=-\dfrac{v}{u}\) and \(u=-u\)
\(2=\dfrac{-(15-u)}{-u}\) \(2 u=+15-u\) \(\Rightarrow 3 u=15\) \( \Rightarrow u = 5\;cm\,;\,\,\,v = 15 - 5 = 10\;cm\) By mirror formula \(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-5}=\dfrac{1-2}{10}\) \( \Rightarrow f = - 10\;cm\)
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364718
An object \(2.5\;cm\) high placed at a distance of \(10\;cm\) from a concave mirror of radius of curvature \(30\;cm\) . The size of the image is
1 \(10.5\;cm\)
2 \(9.2\;cm\)
3 \(7.5\;cm\)
4 \(5.6\;cm\)
Explanation:
\(R = - 30cm \Rightarrow f = - 15cm,\) \(h = + 2.5cm,\,u = - 10cm.\) By mirror formula \(\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{\left( { - 10} \right)}} \Rightarrow v = 30cm\) Also \(\frac{{h'}}{h} = - \frac{v}{u} \Rightarrow \frac{1}{{\left( { + 2.5} \right)}} = \frac{{30}}{{\left( { - 10} \right)}} \Rightarrow I = + 7.5cm\)
364714
A linear object of height \(10\;cm\) is kept in front of a concave mirror of radius of curvature \(15\;cm\), at a distance of \(10\;cm\). The image formed is
364717
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15\,cm,\) the focal length of the mirror must be
1 \( - 10\,cm\)
2 \( - 12\,cm\)
3 \(10/3\,cm\)
4 \(15\,cm\)
Explanation:
Given: \(u+v=15\) \(v=15-u ; m=2=-\dfrac{v}{u}\) and \(u=-u\)
\(2=\dfrac{-(15-u)}{-u}\) \(2 u=+15-u\) \(\Rightarrow 3 u=15\) \( \Rightarrow u = 5\;cm\,;\,\,\,v = 15 - 5 = 10\;cm\) By mirror formula \(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-5}=\dfrac{1-2}{10}\) \( \Rightarrow f = - 10\;cm\)
JEE - 2024
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364718
An object \(2.5\;cm\) high placed at a distance of \(10\;cm\) from a concave mirror of radius of curvature \(30\;cm\) . The size of the image is
1 \(10.5\;cm\)
2 \(9.2\;cm\)
3 \(7.5\;cm\)
4 \(5.6\;cm\)
Explanation:
\(R = - 30cm \Rightarrow f = - 15cm,\) \(h = + 2.5cm,\,u = - 10cm.\) By mirror formula \(\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{\left( { - 10} \right)}} \Rightarrow v = 30cm\) Also \(\frac{{h'}}{h} = - \frac{v}{u} \Rightarrow \frac{1}{{\left( { + 2.5} \right)}} = \frac{{30}}{{\left( { - 10} \right)}} \Rightarrow I = + 7.5cm\)
