Chapter 23, Problem 50AP

(a)

To determine

The focal length of the lens.

Answer to Problem 50AP

The focal length of the lens is $-12.0\text{cm}$.

Explanation of Solution

Given info:

The radius if left side is $-15.0\text{cm}$.

The radius if right side is $10.0\text{cm}$.

The refractive index is $1.50$.

Explanation:

Formula to calculate the focal length is,

$f=\frac{1}{\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$

• $f$ is the focal length
• $n$ is the refractive index
• $R_1$ is the radius of curvature of left side of the lens
• $R_2$ is the radius of curvature of right side of the lens

Substitute $1.50$ for $n$, $-15.0\text{cm}$ for $R_1$ and $10.0\text{cm}$ for $R_2$ to find $f$.

$\begin{array}{c}f=\frac{1}{\left(1.50-1\right)\left(\frac{1}{-15.0\text{cm}}-\frac{1}{\text{10}\text{.0}\text{cm}}\right)}\\ =-12.0\text{cm}\end{array}$

Thus, the focal length of the lens is $-12.0\text{cm}$.

Conclusion:

The focal length of the lens is $-12.0\text{cm}$.

(b)

To determine

The position of the image when the object is at infinity.

Answer to Problem 50AP

The position of the image when the object is at infinity is $-12.0\text{cm}$.

Explanation of Solution

Given info:

The initial object distance is $\propto$.

Explanation:

Formula to calculate the image distance is,

$q=\frac{1}{\left(\frac{1}{f}-\frac{1}{p}\right)}$

• $q$ is the image distance
• $p$ is the object distance

Substitute $\propto$ for $p$ and $-12.0\text{cm}$ for $f$ to find $q$.

$\begin{array}{c}q=\frac{1}{\left(\frac{1}{-12.0\text{cm}}-\frac{1}{\propto }\right)}\\ =-12.0\text{cm}\end{array}$

Thus, the position of the image when the object is at infinity is $-12.0\text{cm}$.

Conclusion:

The position of the image when the object is at infinity is $-12.0\text{cm}$.

(c)

To determine

The position of the image when the object is at a distance $3\left|f\right|$.

Answer to Problem 50AP

The position of the image when the object is at a distance $3\left|f\right|$ is $-9.00\text{cm}$.

Explanation of Solution

Given info:

The initial object distance is $3\left|f\right|$.

Explanation:

Formula to calculate the image distance is,

$q=\frac{1}{\left(\frac{1}{f}-\frac{1}{p}\right)}$

Substitute $3\left|f\right|$ for $p$ and $-12.0\text{cm}$ for $f$ to find $q$.

$\begin{array}{c}q=\frac{1}{\left(\frac{1}{f}-\frac{1}{3\left|f\right|}\right)}\\ =\frac{1}{\left(\frac{1}{-12.0\text{cm}}-\frac{1}{3\left|-12.0\text{cm}\right|}\right)}\\ =-9.00\text{cm}\end{array}$

Thus, the position of the image when the object is at a distance $3\left|f\right|$ is $-9.00\text{cm}$.

Conclusion:

The position of the image when the object is at a distance $3\left|f\right|$ is $-9.00\text{cm}$.

(d)

To determine

The position of the image when the object is at a distance $\left|f\right|$.

Answer to Problem 50AP

The position of the image when the object is at a distance $\left|f\right|$ is $-6.00\text{cm}$.

Explanation of Solution

Given info:

The initial object distance is $\left|f\right|$.

Explanation:

Formula to calculate the image distance is,

$q=\frac{1}{\left(\frac{1}{f}-\frac{1}{p}\right)}$

Substitute $\left|f\right|$ for $p$ and $-12.0\text{cm}$ for $f$ to find $q$.

$\begin{array}{c}p=\frac{1}{\left(\frac{1}{f}-\frac{1}{\left|f\right|}\right)}\\ =\frac{1}{\left(\frac{1}{-12.0\text{cm}}-\frac{1}{\left|-12.0\text{cm}\right|}\right)}\\ =-6.00\text{cm}\end{array}$

Thus, the position of the image when the object is at a distance $\left|f\right|$ is $-6.00\text{cm}$.

Conclusion:

The position of the image when the object is at a distance $\left|f\right|$ is $-6.00\text{cm}$.

(e)

To determine

The position of the image when the object is at a distance $\left|f\right|/2$.

Answer to Problem 50AP

The position of the image when the object is at a distance $\left|f\right|/2$ is $-4.00\text{cm}$

Explanation of Solution

Given info:

The initial object distance is $\left|f\right|/2$.

Explanation:

Formula to calculate the image distance is,

$q=\frac{1}{\left(\frac{1}{f}-\frac{1}{p}\right)}$

Substitute $\left|f\right|/2$ for $p$ and $-12.0\text{cm}$ for $f$ to find $q$.

$\begin{array}{c}q=\frac{1}{\left(\frac{1}{f}-\frac{1}{\left|f\right|/2}\right)}\\ =\frac{1}{\left(\frac{1}{-12.0\text{cm}}-\frac{1}{\left|-12.0\text{cm}\right|/2}\right)}\\ =-4.00\text{cm}\end{array}$

Thus, the position of the image when the object is at a distance $\left|f\right|/2$ is $-4.00\text{cm}$.

Conclusion:

The position of the image when the object is at a distance $\left|f\right|/2$ is $-4.00\text{cm}$.