364706
An object is placed at \(20\;cm\) in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?
1 \(10\;cm\)
2 \(15\;cm\)
3 \(7.5\;cm\)
4 \(6.6\;cm\)
Explanation:
As image is real, magnification \(m = - \frac{v}{u} = - 3 \Rightarrow v = 3u\) Here, \(u = - 20{\mkern 1mu} \;cm.\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} v = - 60\;cm.\) According to mirror formula, \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) \( \Rightarrow \frac{1}{f} = \frac{1}{{ - 60}} + \frac{1}{{ - 20}} = - \frac{4}{{60}}\) \( \Rightarrow f = - 15\;{\mkern 1mu} cm\; \Rightarrow \;\left| f \right| = 15\;{\mkern 1mu} cm\)
KCET - 2014
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364707
A square \({A B C D}\) of side \(1\,mm\) is kept at distance \(15\,cm\) infront of the concave mirror as shown in the figure. The focal length of the mirror is \(10\,cm\). The length of the perimeter of its image will be (nearly):
1 \(8\,mm\)
2 \(2\,mm\)
3 \(12\,mm\)
4 \(6\,mm\)
Explanation:
By mirror formula, \({\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{15}=\dfrac{5}{-150}=\dfrac{1}{-30}}\) So, \({v=-30 {~cm}}\) Let \({A^{\prime} B^{\prime}}\) and \({C^{\prime} D^{\prime}}\) be image of \({A B}\) and \({C D}\) respectively. \({m=-\dfrac{v}{u}=-2}\) \({\therefore A^{\prime} B^{\prime}=C^{\prime} D^{\prime}=2 \times 1=2 {~mm}}\) Let \({B^{\prime} C^{\prime}}\) and \({A^{\prime} D^{\prime}}\) be image of \({B C}\) and \({A D}\) respectively. \(\frac{{{B^\prime }{C^\prime }}}{{BC}} = \frac{{{A^\prime }{D^\prime }}}{{AD}} = \frac{{{v^2}}}{{{u^2}}} = 4\) \( \Rightarrow {B^\prime }{C^\prime } = {A^\prime }{D^\prime } = 4\;mm\) \({\therefore}\) Length \({=2+2+4+4=12 {~mm}}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364708
A short linear object of length \(b\) lies along the axis of a concave mirror. The size of the image is equal to
1 \(b\left( {\frac{f}{{u - f}}} \right)\)
2 \(b{\left( {\frac{{u - f}}{f}} \right)^{1/2}}\)
3 \(b{\left( {\frac{f}{{f - u}}} \right)^2}\)
4 \(b\left( {\frac{{u - f}}{f}} \right)\)
Explanation:
For a spherical mirror \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{{ - du}}{{{u^2}}} - \frac{{dv}}{{{v^2}}} = 0\) \(dv = - {\left( {\frac{v}{u}} \right)^2}du = - {m^2}du\) Where \(du\) and \(dv\) are lenghts of small object and its image \(dv = - {\left( {\frac{f}{{f - u}}} \right)^2}b\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364709
With a concave mirror, an object is placed at a distance \({x_1}\) from the principle focus, on the principle axis. The image is formed at a distance \({x_2}\) from the principle focus. The focal length of the mirror is
364706
An object is placed at \(20\;cm\) in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?
1 \(10\;cm\)
2 \(15\;cm\)
3 \(7.5\;cm\)
4 \(6.6\;cm\)
Explanation:
As image is real, magnification \(m = - \frac{v}{u} = - 3 \Rightarrow v = 3u\) Here, \(u = - 20{\mkern 1mu} \;cm.\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} v = - 60\;cm.\) According to mirror formula, \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) \( \Rightarrow \frac{1}{f} = \frac{1}{{ - 60}} + \frac{1}{{ - 20}} = - \frac{4}{{60}}\) \( \Rightarrow f = - 15\;{\mkern 1mu} cm\; \Rightarrow \;\left| f \right| = 15\;{\mkern 1mu} cm\)
KCET - 2014
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364707
A square \({A B C D}\) of side \(1\,mm\) is kept at distance \(15\,cm\) infront of the concave mirror as shown in the figure. The focal length of the mirror is \(10\,cm\). The length of the perimeter of its image will be (nearly):
1 \(8\,mm\)
2 \(2\,mm\)
3 \(12\,mm\)
4 \(6\,mm\)
Explanation:
By mirror formula, \({\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{15}=\dfrac{5}{-150}=\dfrac{1}{-30}}\) So, \({v=-30 {~cm}}\) Let \({A^{\prime} B^{\prime}}\) and \({C^{\prime} D^{\prime}}\) be image of \({A B}\) and \({C D}\) respectively. \({m=-\dfrac{v}{u}=-2}\) \({\therefore A^{\prime} B^{\prime}=C^{\prime} D^{\prime}=2 \times 1=2 {~mm}}\) Let \({B^{\prime} C^{\prime}}\) and \({A^{\prime} D^{\prime}}\) be image of \({B C}\) and \({A D}\) respectively. \(\frac{{{B^\prime }{C^\prime }}}{{BC}} = \frac{{{A^\prime }{D^\prime }}}{{AD}} = \frac{{{v^2}}}{{{u^2}}} = 4\) \( \Rightarrow {B^\prime }{C^\prime } = {A^\prime }{D^\prime } = 4\;mm\) \({\therefore}\) Length \({=2+2+4+4=12 {~mm}}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364708
A short linear object of length \(b\) lies along the axis of a concave mirror. The size of the image is equal to
1 \(b\left( {\frac{f}{{u - f}}} \right)\)
2 \(b{\left( {\frac{{u - f}}{f}} \right)^{1/2}}\)
3 \(b{\left( {\frac{f}{{f - u}}} \right)^2}\)
4 \(b\left( {\frac{{u - f}}{f}} \right)\)
Explanation:
For a spherical mirror \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{{ - du}}{{{u^2}}} - \frac{{dv}}{{{v^2}}} = 0\) \(dv = - {\left( {\frac{v}{u}} \right)^2}du = - {m^2}du\) Where \(du\) and \(dv\) are lenghts of small object and its image \(dv = - {\left( {\frac{f}{{f - u}}} \right)^2}b\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364709
With a concave mirror, an object is placed at a distance \({x_1}\) from the principle focus, on the principle axis. The image is formed at a distance \({x_2}\) from the principle focus. The focal length of the mirror is
364706
An object is placed at \(20\;cm\) in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?
1 \(10\;cm\)
2 \(15\;cm\)
3 \(7.5\;cm\)
4 \(6.6\;cm\)
Explanation:
As image is real, magnification \(m = - \frac{v}{u} = - 3 \Rightarrow v = 3u\) Here, \(u = - 20{\mkern 1mu} \;cm.\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} v = - 60\;cm.\) According to mirror formula, \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) \( \Rightarrow \frac{1}{f} = \frac{1}{{ - 60}} + \frac{1}{{ - 20}} = - \frac{4}{{60}}\) \( \Rightarrow f = - 15\;{\mkern 1mu} cm\; \Rightarrow \;\left| f \right| = 15\;{\mkern 1mu} cm\)
KCET - 2014
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364707
A square \({A B C D}\) of side \(1\,mm\) is kept at distance \(15\,cm\) infront of the concave mirror as shown in the figure. The focal length of the mirror is \(10\,cm\). The length of the perimeter of its image will be (nearly):
1 \(8\,mm\)
2 \(2\,mm\)
3 \(12\,mm\)
4 \(6\,mm\)
Explanation:
By mirror formula, \({\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{15}=\dfrac{5}{-150}=\dfrac{1}{-30}}\) So, \({v=-30 {~cm}}\) Let \({A^{\prime} B^{\prime}}\) and \({C^{\prime} D^{\prime}}\) be image of \({A B}\) and \({C D}\) respectively. \({m=-\dfrac{v}{u}=-2}\) \({\therefore A^{\prime} B^{\prime}=C^{\prime} D^{\prime}=2 \times 1=2 {~mm}}\) Let \({B^{\prime} C^{\prime}}\) and \({A^{\prime} D^{\prime}}\) be image of \({B C}\) and \({A D}\) respectively. \(\frac{{{B^\prime }{C^\prime }}}{{BC}} = \frac{{{A^\prime }{D^\prime }}}{{AD}} = \frac{{{v^2}}}{{{u^2}}} = 4\) \( \Rightarrow {B^\prime }{C^\prime } = {A^\prime }{D^\prime } = 4\;mm\) \({\therefore}\) Length \({=2+2+4+4=12 {~mm}}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364708
A short linear object of length \(b\) lies along the axis of a concave mirror. The size of the image is equal to
1 \(b\left( {\frac{f}{{u - f}}} \right)\)
2 \(b{\left( {\frac{{u - f}}{f}} \right)^{1/2}}\)
3 \(b{\left( {\frac{f}{{f - u}}} \right)^2}\)
4 \(b\left( {\frac{{u - f}}{f}} \right)\)
Explanation:
For a spherical mirror \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{{ - du}}{{{u^2}}} - \frac{{dv}}{{{v^2}}} = 0\) \(dv = - {\left( {\frac{v}{u}} \right)^2}du = - {m^2}du\) Where \(du\) and \(dv\) are lenghts of small object and its image \(dv = - {\left( {\frac{f}{{f - u}}} \right)^2}b\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364709
With a concave mirror, an object is placed at a distance \({x_1}\) from the principle focus, on the principle axis. The image is formed at a distance \({x_2}\) from the principle focus. The focal length of the mirror is
364706
An object is placed at \(20\;cm\) in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?
1 \(10\;cm\)
2 \(15\;cm\)
3 \(7.5\;cm\)
4 \(6.6\;cm\)
Explanation:
As image is real, magnification \(m = - \frac{v}{u} = - 3 \Rightarrow v = 3u\) Here, \(u = - 20{\mkern 1mu} \;cm.\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} v = - 60\;cm.\) According to mirror formula, \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) \( \Rightarrow \frac{1}{f} = \frac{1}{{ - 60}} + \frac{1}{{ - 20}} = - \frac{4}{{60}}\) \( \Rightarrow f = - 15\;{\mkern 1mu} cm\; \Rightarrow \;\left| f \right| = 15\;{\mkern 1mu} cm\)
KCET - 2014
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364707
A square \({A B C D}\) of side \(1\,mm\) is kept at distance \(15\,cm\) infront of the concave mirror as shown in the figure. The focal length of the mirror is \(10\,cm\). The length of the perimeter of its image will be (nearly):
1 \(8\,mm\)
2 \(2\,mm\)
3 \(12\,mm\)
4 \(6\,mm\)
Explanation:
By mirror formula, \({\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{15}=\dfrac{5}{-150}=\dfrac{1}{-30}}\) So, \({v=-30 {~cm}}\) Let \({A^{\prime} B^{\prime}}\) and \({C^{\prime} D^{\prime}}\) be image of \({A B}\) and \({C D}\) respectively. \({m=-\dfrac{v}{u}=-2}\) \({\therefore A^{\prime} B^{\prime}=C^{\prime} D^{\prime}=2 \times 1=2 {~mm}}\) Let \({B^{\prime} C^{\prime}}\) and \({A^{\prime} D^{\prime}}\) be image of \({B C}\) and \({A D}\) respectively. \(\frac{{{B^\prime }{C^\prime }}}{{BC}} = \frac{{{A^\prime }{D^\prime }}}{{AD}} = \frac{{{v^2}}}{{{u^2}}} = 4\) \( \Rightarrow {B^\prime }{C^\prime } = {A^\prime }{D^\prime } = 4\;mm\) \({\therefore}\) Length \({=2+2+4+4=12 {~mm}}\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364708
A short linear object of length \(b\) lies along the axis of a concave mirror. The size of the image is equal to
1 \(b\left( {\frac{f}{{u - f}}} \right)\)
2 \(b{\left( {\frac{{u - f}}{f}} \right)^{1/2}}\)
3 \(b{\left( {\frac{f}{{f - u}}} \right)^2}\)
4 \(b\left( {\frac{{u - f}}{f}} \right)\)
Explanation:
For a spherical mirror \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{{ - du}}{{{u^2}}} - \frac{{dv}}{{{v^2}}} = 0\) \(dv = - {\left( {\frac{v}{u}} \right)^2}du = - {m^2}du\) Where \(du\) and \(dv\) are lenghts of small object and its image \(dv = - {\left( {\frac{f}{{f - u}}} \right)^2}b\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364709
With a concave mirror, an object is placed at a distance \({x_1}\) from the principle focus, on the principle axis. The image is formed at a distance \({x_2}\) from the principle focus. The focal length of the mirror is
