Chapter 36, Problem 26SP

A spherical concave mirror has a radius of curvature of -400 cm. An object 2.00 cm tall is on the central axis 400 cm in front of the mirror. (a) Determine the focal length. (b) Locate the image. (c) Describe the image. (d) Determine the magnification. [Hint: Check out Fig. 36-5.]

To determine

The focal length of a $2.00\text{ cm}$ tall object that is located $400\text{ cm}$ in front of a concave mirror, which has a radius of curvature of $-400\text{ cm}$.

Answer to Problem 26SP

Solution:

$+200\text{ cm}$

Explanation of Solution

Given data:

The radius of curvature is $-400\text{ cm}$.

The height of the object is $2.00\text{ cm}$.

The distance of the object from the concave mirror is $400\text{ cm}$.

Formula used:

The thin mirror equation is written as,

$\frac{1}{f}=-\frac{2}{R}$

Here, $f$ is the focal length of the mirror and $R$ is the radius of curvature of the mirror.

Sign convention:

If R is negative, the centre of curvature is to the left of the mirror, and the mirror is concave.

If R is positive, the centre of curvature is to the right of the mirror, and the mirror is convex.

If f is positive, the mirror is concave.

If f is negative, the mirror is convex.

Explanation:

Recall the expression forthin mirror.

$\frac{1}{f}=-\frac{2}{R}$

Solve for $f$.

$f=-\frac{R}{2}$

Substitute $-400\text{ cm}$ for $R$

$\begin{array}{c}f=-\frac{\left(-400\text{ cm}\right)}{2}\\ =+200\text{ cm}\end{array}$

The positive sign indicates that the mirror is concave.

Conclusion:

Therefore, the focal length of the mirror is $+200\text{ cm}$.

To determine

The location of the image, when a $2.00\text{ cm}$ tall object is located $400\text{ cm}$ in front of a concave mirror, which has a radius of curvature of $-400\text{ cm}$.

Answer to Problem 26SP

Solution:

The real image is $400\text{ cm}$ to the left of the mirror.

Explanation of Solution

Given data:

The radius of curvature is $-400\text{ cm}$.

The height of the object is $2.00\text{ cm}$.

The distance of the object from the concave mirror is $400\text{ cm}$.

From previous part, the focal length of the given concave mirror is $+200\text{ cm}$.

Formula used:

The thin mirror equation is written as,

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Here, $f$ is the focal length of the mirror, $s_o$ is the object distance from the mirror, and $s_i$ is the image distance from the concave mirror.

Sign convention:

If $s_o$ is positive, the object is in the front (that is to the left) of the mirror.

If $s_i$ is positive, the image is real (that isin front or to the left of the mirror).

If $s_i$ is negative, the image is virtual (that is behind or to the right of the mirror).

If f is positive, the mirror is concave.

If f is negative, the mirror is convex.

Explanation:

Consider the expression forthinmirror.

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Understand that the object is placed at a distance of $2f$ from the mirror.

Substitute $400$ for $s_o$ and $200\text{ cm}$ for $f$

$\frac{1}{\left(200\text{ cm}\right)}=\frac{1}{400\text{ cm}}+\frac{1}{s_i}$

Solve for $s_i$.

$s_i=+400\text{ cm}$

The positive sign indicates that the image is real.

Conclusion:

Therefore, the location of t hereal image is $400\text{ cm}$ to the left of the mirror.

To determine

The nature of the image, when a $2.00\text{ cm}$ tall object is located $400\text{ cm}$ in front of a concave mirror, which has a radius of curvature of $-400\text{ cm}$.

Answer to Problem 26SP

Solution:

The image is r eal, inverted, and of the same size as the object.

Explanation of Solution

Introduction:

From table 36-1 in the textbook, it is clear that when the object is at a distance of $2f$ $\left(s_o=2f\right)$ from the concave mirror, the image is formed at a distance of $2f$ $\left(s_i=2f\right)$.

Explanation:

Draw the diagram ofa concave mirror when the object is placed at a distance of $2f$.

From the above figure, it is clear that when the object is placed at a distance of $2f$, or at the centre of curvature $\left(\text{C}\right)$ distance from the mirror, the image is formed real, inverted and of the same size as the object on the centre of curvature.

Conclusion:

Therefore, the image formed by the concave mirror is real, inverted , and of the same size as the object.

To determine

The magnification, when a $2.00\text{ cm}$ tall object is located $400\text{ cm}$ in front of a concave mirror, which has a radius of curvature of $-400\text{ cm}$.

Answer to Problem 26SP

Solution:

$-1$

Explanation of Solution

Given data:

The radius of curvature is $-400\text{ cm}$.

The height of the object is $2.00\text{ cm}$.

The distance of the object from the concave mirror is $400\text{ cm}$.

From part (a), the focal length of the given concave mirror is $+200\text{ cm}$.

Formula used:

The formula for the magnification of the mirror is:

$M=\frac{y_i}{y_o}=-\frac{s_i}{s_o}$

Here, $y_i$ is the height of the image and $y_o$ is the height of the object.

Explanation:

Consider the expression for the magnification of the mirror.

$M=-\frac{s_i}{s_o}$

Obtain the value of $s_i$ from part (b).

$s_i=400\text{ cm}$

Substitute $400\text{ cm}$ for $s_o$ and $400\text{ cm}$ for $s_i$

$\begin{array}{c}M=-\frac{400\text{ cm}}{400\text{ cm}}\\ =-1\end{array}$

The negative sign indicates that the image is inverted.

Conclusion:

Therefore, the magnification is $-1$.