364701
A convex mirror of focal length f forms an image which is \(\frac{1}{n}\) times the object. The distance of the object from the mirror is
1 \(\left( {n + 1} \right)f\)
2 \(\left( {n - 1} \right)f\)
3 \(\left( {n + \frac{1}{2}} \right)f\)
4 \(\left( {\frac{{n + 1}}{n}} \right)f\)
Explanation:
\(m = + \frac{1}{n} = - \frac{v}{u} \Rightarrow v = - \frac{u}{n}\) By using mirror formula \(\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}\) \( \Rightarrow u = - \left( {n - 1} \right)f\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364702
An object is placed at a distance equal to focal length of convex mirror. If the focal length of the mirror be \(f\), then the distance of the image from the pole of the mirror is
1 less than \(f\)
2 equal to \(f\)
3 more than \(f\)
4 infinity
Explanation:
Image formed by convex mirror is shown in figure. \(u = - f\) By mirror formula, \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2}\) \(i.e.\) the distance of the image formed is less than \(f.\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364703
A convex mirror of radius of curvatute \(30\,cm\) forms an image that is half the size of the object. The object distance is
364704
A square wire of side \(4.8 {~cm}\) is placed \(90 {~cm}\) away from a concave mirror of focal length \(18 {~cm}\) as shown in the figure (not to the scale). What is the area enclosed by the image of the wire in \({cm}^{2}\) ?
364705
A spherical mirror forms an image of magnification \(m = \pm 3.\) The object distance, if focal length of mirror is \(24{\rm{ }}cm,{\rm{ }}\) may be
364701
A convex mirror of focal length f forms an image which is \(\frac{1}{n}\) times the object. The distance of the object from the mirror is
1 \(\left( {n + 1} \right)f\)
2 \(\left( {n - 1} \right)f\)
3 \(\left( {n + \frac{1}{2}} \right)f\)
4 \(\left( {\frac{{n + 1}}{n}} \right)f\)
Explanation:
\(m = + \frac{1}{n} = - \frac{v}{u} \Rightarrow v = - \frac{u}{n}\) By using mirror formula \(\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}\) \( \Rightarrow u = - \left( {n - 1} \right)f\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364702
An object is placed at a distance equal to focal length of convex mirror. If the focal length of the mirror be \(f\), then the distance of the image from the pole of the mirror is
1 less than \(f\)
2 equal to \(f\)
3 more than \(f\)
4 infinity
Explanation:
Image formed by convex mirror is shown in figure. \(u = - f\) By mirror formula, \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2}\) \(i.e.\) the distance of the image formed is less than \(f.\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364703
A convex mirror of radius of curvatute \(30\,cm\) forms an image that is half the size of the object. The object distance is
364704
A square wire of side \(4.8 {~cm}\) is placed \(90 {~cm}\) away from a concave mirror of focal length \(18 {~cm}\) as shown in the figure (not to the scale). What is the area enclosed by the image of the wire in \({cm}^{2}\) ?
364705
A spherical mirror forms an image of magnification \(m = \pm 3.\) The object distance, if focal length of mirror is \(24{\rm{ }}cm,{\rm{ }}\) may be
364701
A convex mirror of focal length f forms an image which is \(\frac{1}{n}\) times the object. The distance of the object from the mirror is
1 \(\left( {n + 1} \right)f\)
2 \(\left( {n - 1} \right)f\)
3 \(\left( {n + \frac{1}{2}} \right)f\)
4 \(\left( {\frac{{n + 1}}{n}} \right)f\)
Explanation:
\(m = + \frac{1}{n} = - \frac{v}{u} \Rightarrow v = - \frac{u}{n}\) By using mirror formula \(\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}\) \( \Rightarrow u = - \left( {n - 1} \right)f\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364702
An object is placed at a distance equal to focal length of convex mirror. If the focal length of the mirror be \(f\), then the distance of the image from the pole of the mirror is
1 less than \(f\)
2 equal to \(f\)
3 more than \(f\)
4 infinity
Explanation:
Image formed by convex mirror is shown in figure. \(u = - f\) By mirror formula, \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2}\) \(i.e.\) the distance of the image formed is less than \(f.\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364703
A convex mirror of radius of curvatute \(30\,cm\) forms an image that is half the size of the object. The object distance is
364704
A square wire of side \(4.8 {~cm}\) is placed \(90 {~cm}\) away from a concave mirror of focal length \(18 {~cm}\) as shown in the figure (not to the scale). What is the area enclosed by the image of the wire in \({cm}^{2}\) ?
364705
A spherical mirror forms an image of magnification \(m = \pm 3.\) The object distance, if focal length of mirror is \(24{\rm{ }}cm,{\rm{ }}\) may be
364701
A convex mirror of focal length f forms an image which is \(\frac{1}{n}\) times the object. The distance of the object from the mirror is
1 \(\left( {n + 1} \right)f\)
2 \(\left( {n - 1} \right)f\)
3 \(\left( {n + \frac{1}{2}} \right)f\)
4 \(\left( {\frac{{n + 1}}{n}} \right)f\)
Explanation:
\(m = + \frac{1}{n} = - \frac{v}{u} \Rightarrow v = - \frac{u}{n}\) By using mirror formula \(\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}\) \( \Rightarrow u = - \left( {n - 1} \right)f\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364702
An object is placed at a distance equal to focal length of convex mirror. If the focal length of the mirror be \(f\), then the distance of the image from the pole of the mirror is
1 less than \(f\)
2 equal to \(f\)
3 more than \(f\)
4 infinity
Explanation:
Image formed by convex mirror is shown in figure. \(u = - f\) By mirror formula, \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2}\) \(i.e.\) the distance of the image formed is less than \(f.\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364703
A convex mirror of radius of curvatute \(30\,cm\) forms an image that is half the size of the object. The object distance is
364704
A square wire of side \(4.8 {~cm}\) is placed \(90 {~cm}\) away from a concave mirror of focal length \(18 {~cm}\) as shown in the figure (not to the scale). What is the area enclosed by the image of the wire in \({cm}^{2}\) ?
364705
A spherical mirror forms an image of magnification \(m = \pm 3.\) The object distance, if focal length of mirror is \(24{\rm{ }}cm,{\rm{ }}\) may be
364701
A convex mirror of focal length f forms an image which is \(\frac{1}{n}\) times the object. The distance of the object from the mirror is
1 \(\left( {n + 1} \right)f\)
2 \(\left( {n - 1} \right)f\)
3 \(\left( {n + \frac{1}{2}} \right)f\)
4 \(\left( {\frac{{n + 1}}{n}} \right)f\)
Explanation:
\(m = + \frac{1}{n} = - \frac{v}{u} \Rightarrow v = - \frac{u}{n}\) By using mirror formula \(\frac{1}{f} = \frac{1}{{ - \frac{u}{n}}} + \frac{1}{u}\) \( \Rightarrow u = - \left( {n - 1} \right)f\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364702
An object is placed at a distance equal to focal length of convex mirror. If the focal length of the mirror be \(f\), then the distance of the image from the pole of the mirror is
1 less than \(f\)
2 equal to \(f\)
3 more than \(f\)
4 infinity
Explanation:
Image formed by convex mirror is shown in figure. \(u = - f\) By mirror formula, \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2}\) \(i.e.\) the distance of the image formed is less than \(f.\)
PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS
364703
A convex mirror of radius of curvatute \(30\,cm\) forms an image that is half the size of the object. The object distance is
364704
A square wire of side \(4.8 {~cm}\) is placed \(90 {~cm}\) away from a concave mirror of focal length \(18 {~cm}\) as shown in the figure (not to the scale). What is the area enclosed by the image of the wire in \({cm}^{2}\) ?
364705
A spherical mirror forms an image of magnification \(m = \pm 3.\) The object distance, if focal length of mirror is \(24{\rm{ }}cm,{\rm{ }}\) may be
