Physics: An object that is 2.00 cm high is placed 10.0 cm in front of a concave mirror with a radius of curvature of 40.0 cm. Find the magnification and location of the corresponding image in relation to the mirror’s surface.

When a substance changes from a liquid to a gas, we say that it vaporizes. Vaporization is the change from a liquid state to a gaseous state. As a substance is heated, its particles begin to move faster and faster.Explanation:...

An object that is 2.00 cm high is placed 10.0 cm

Ответ:

ivanf30

01.07.2020

Physics

The magnification is 1, that is the image is upright and the same size as the object

The location of the image is behind the mirror (virtual image)

Explanation:

Here we have the mirror equation presented as follows;

$\frac{1}{d_o}+ \frac{1}{d_i} = \frac{1}{f}$

$m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$

Where:

d_o = Distance of the object from the mirror = 10.0 cm

d_i = Distance of the image from the mirror

R = Radius of curvature of the mirror = 40.0 cm

f = Focal length of the mirror = 1/2 × R = 1/2 × 40 = 20 cm

m = Magnification of the mirror

h_o = Height of the object = 2.00 cm

h_i = Height of the image

Plugging in the values, we have

$\frac{1}{10}+ \frac{1}{d_i} = \frac{1}{20}$

$\frac{1}{d_i} = \frac{1}{20} - \frac{1}{10} = - \frac{1}{10}$

∴ d_i = -10 cm which means a virtual image, behind the mirror

The magnification is found and presented as follows;

$m = -\frac{d_i}{d_o} = -\frac{-10}{10} = 1$

$h_i = -\frac{d_i}{d_o} \times h_o = -\frac{-10}{10} \times 2 = 2 \, cm$

Therefore, the image is upright and the same size as the object.

0,0(0 оценок)

An object that is 2.00 cm high is placed 10.0 cm in front of a concave mirror with a radius of curvature of 40.0 cm. Find the magnification and location of the corresponding image in relation to the mirror’s surface.

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question