Assume that a certain concave spherical mirror has a focal length of 10.0 cm.
(a)Locate the image and find the magnification for an object distance of 25.0 cm. Determine whether the image is real or virtual, inverted or upright, and larger or smaller. Do the same for object distances of
(b)10.0 cm and(c)5.00 cm
The correct answer and explanation is :
To solve this problem involving a concave mirror, we will use the mirror equation and magnification formula. Let’s begin by recalling these equations:
Mirror Equation:
[
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
]
where:
- ( f ) is the focal length of the mirror,
- ( d_o ) is the object distance,
- ( d_i ) is the image distance.
Magnification Equation:
[
M = -\frac{d_i}{d_o}
]
where:
- ( M ) is the magnification,
- ( d_i ) is the image distance,
- ( d_o ) is the object distance.
For a concave mirror, the focal length (( f )) is positive, and the image characteristics depend on the object distance (( d_o )).
Given:
- Focal length of the mirror, ( f = 10.0 \, \text{cm} ).
We will now solve the problem for three different object distances.
(a) For object distance ( d_o = 25.0 \, \text{cm} ):
- Using the mirror equation:
[
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
]
Substitute ( f = 10.0 \, \text{cm} ) and ( d_o = 25.0 \, \text{cm} ):
[
\frac{1}{10} = \frac{1}{25} + \frac{1}{d_i}
]
Solve for ( d_i ):
[
\frac{1}{d_i} = \frac{1}{10} – \frac{1}{25} = \frac{5 – 2}{50} = \frac{3}{50}
]
[
d_i = \frac{50}{3} \approx 16.67 \, \text{cm}
]
The image distance is positive, meaning the image is real.
- Finding magnification:
[
M = -\frac{d_i}{d_o} = -\frac{16.67}{25.0} \approx -0.667
]
The magnification is negative, indicating that the image is inverted. Since the absolute value of ( M ) is less than 1, the image is smaller than the object.
(b) For object distance ( d_o = 10.0 \, \text{cm} ):
- Using the mirror equation:
[
\frac{1}{10} = \frac{1}{10} + \frac{1}{d_i}
]
Solve for ( d_i ):
[
\frac{1}{d_i} = \frac{1}{10} – \frac{1}{10} = 0
]
This means ( d_i ) is infinite, implying that the image is formed at infinity.
- Magnification:
Since the image is at infinity, the magnification is also undefined or effectively very large. The image is real and highly magnified.
(c) For object distance ( d_o = 5.00 \, \text{cm} ):
- Using the mirror equation:
[
\frac{1}{10} = \frac{1}{5.00} + \frac{1}{d_i}
]
Solve for ( d_i ):
[
\frac{1}{d_i} = \frac{1}{10} – \frac{1}{5} = \frac{1 – 2}{10} = -\frac{1}{10}
]
[
d_i = -10.0 \, \text{cm}
]
The image distance is negative, meaning the image is virtual.
- Magnification:
[
M = -\frac{d_i}{d_o} = -\frac{-10.0}{5.00} = 2.0
]
The magnification is positive and greater than 1, indicating that the image is upright and larger than the object.
Summary of Results:
- (a) ( d_o = 25.0 \, \text{cm} ): Real, inverted, and smaller image. Magnification = (-0.667), image distance ( d_i = 16.67 \, \text{cm} ).
- (b) ( d_o = 10.0 \, \text{cm} ): Real, highly magnified image at infinity.
- (c) ( d_o = 5.00 \, \text{cm} ): Virtual, upright, and larger image. Magnification = ( 2.0 ), image distance ( d_i = -10.0 \, \text{cm} ).
The image’s nature (real/virtual), orientation (inverted/upright), and size (larger/smaller) depend critically on the object distance relative to the focal length.