Question

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird located \(100 . \mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?

Short answer

Answer: No, while the image is indeed inverted, it is only 0.0408 cm tall, which is much smaller than 1 cm.

Step-by-step solution

Identify the mirror type and the variables given

The focal length of the mirror is \(40.0 cm\), it is a concave mirror because these types of mirrors are used to project images. The object, a bird, is located \(100 m = 10000 cm\) away from the mirror and it is \(10.0 cm\) tall. The classmate's claim is that the image should be \(.0 cm\) tall and inverted.

Use the mirror formula to find the image distance

The mirror formula is given by the equation \(\frac{1}{f} = \frac{1}{ d_o} +\frac{1}{ d_i}\). Here, \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. We should find the image distance using the given values. Plug in the given values: \(f = 40.0cm\) and \(d_o = 10000cm\) to calculate \(d_i\): \(\frac{1}{40} = \frac{1}{10000} + \frac{1}{d_i}\) Now, solve for \(d_i\): \(\frac{1}{d_i} = \frac{1}{40} - \frac{1}{10000}\) \(d_i=\frac{1}{(\frac{1}{40} - \frac{1}{10000})}\) Calculating this, we get: \(d_i=40.8cm\)

Calculate the magnification

The magnification \(M\) of a mirror is given by the formula \(M = -\frac{d_i}{ d_o}\), where \(d_i\) is the image distance and \(d_o\) is the object distance. Plug in the values we have: \(d_i = 40.8cm\) and \(d_o = 10000cm\): \(M = -\frac{40.8}{10000} = -0.00408\)

Find the image height and analyze the result

To find the image height, we can use the magnification formula \(M = \frac{h_i}{h_o}\), where \(h_i\) is the image height and \(h_o\) is the object height. We have \(M = -0.00408\) and \(h_o = 10.0cm\). Rearranging the formula to find \(h_i\): \(h_i= M \cdot h_o\) Plug in the values: \(M=-0.00408\) and \(h_o=10.0cm\): \(h_i=-0.00408 \cdot 10.0\) \(h_i = -0.0408cm\) The negative value of \(h_i\) indicates that the image is indeed inverted, as the classmate claimed. However, the image height is only \(0.0408 cm\), which is smaller than the claimed \(1 cm\). In conclusion, the claim that the image is no less than \(1.00cm\) tall and inverted is not true; while the image is indeed inverted, it is only \(0.0408 cm\) tall, which is much smaller than \(1 cm\).