Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length $f$ of the lens to the object distance $p_1$.

Answer to Problem 90AP

The equation relating the focal length $f$ of the lens to the object distance $p_1$ is $\frac{1}{f}=\frac{1}{p_1}+\frac{1}{1.50\text{m}-p_1}$.

Explanation of Solution

Given info: The distance between the candle and the wall is $1.50\text{m}$. The distance by which the lens is moved is $90.0\text{cm}$. The focal length of the lens is $f$, the object distance is $p_1$.

Write the expression of lens equation.

$\frac{1}{f}=\frac{1}{p_1}+\frac{1}{q_1}$ (1)

Here,

$q_1$ is the distance of image from the lens.

The sum of image and object distance is,

$\begin{array}{c}{p}_1+q_1=1.50\text{m}\\ q_1=1.50\text{m}-p_1\end{array}$

Substitute $1.50\text{m}-p_1$ for $q_1$ in equation (1).

$\frac{1}{f}=\frac{1}{p_1}+\frac{1}{1.50\text{m}-p_1}$ . (2)

Conclusion:

Therefore, the equation relating the focal length $f$ of the lens to the object distance $p_1$ is $\frac{1}{f}=\frac{1}{p_1}+\frac{1}{1.50\text{m}-p_1}$.

(b)

To determine

The equation relating the focal length $f$ of the lens to the object distance $p_1$.

Answer to Problem 90AP

The equation relating the focal length $f$ of the lens to the object distance $p_1$ is $\frac{1}{f}=\frac{1}{p_1+0.900\text{m}}+\frac{1}{\text{0}\text{.600}\text{m}-p_1}$.

Explanation of Solution

Given info: The distance between the candle and the wall is $1.50\text{m}$. The distance by which the lens is moved is $90.0\text{cm}$. The focal length of the lens is $f$, the object distance is $p_1$.

Write the expression of lens equation.

$\frac{1}{f}=\frac{1}{p_2}+\frac{1}{q_2}$ (3)

Here,

$p_2$ is the final object distance.

$q_2$ is the final image distance.

The final distance of object is,

$\begin{array}{c}{p}_2=p_1+90\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =p_1+0.900\text{m}\end{array}$

The final distance of image is,

$\begin{array}{c}{q}_2=q_1-90\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =q_1-0.900\text{m}\end{array}$

Substitute $1.50\text{m}-p_1$ for $q_1$ in above equation.

$\begin{array}{c}{q}_2=1.50\text{m}-p_1-0.900\text{m}\\ \text{=0}\text{.600}\text{m}-p_1\end{array}$

Substitute $\text{0}\text{.600}\text{m}-p_1$ for $q_2$ and $p_1+0.900\text{m}$ for $p_2$ in equation (3).

$\frac{1}{f}=\frac{1}{p_1+0.900\text{m}}+\frac{1}{\text{0}\text{.600}\text{m}-p_1}$ (4)

Thus, the equation relating the focal length $f$ of the lens to the object distance $p_1$ is $\frac{1}{f}=\frac{1}{p_1+0.900\text{m}}+\frac{1}{\text{0}\text{.600}\text{m}-p_1}$.

Conclusion:

Therefore, the equation relating the focal length $f$ of the lens to the object distance $p_1$ is $\frac{1}{f}=\frac{1}{p_1+0.900\text{m}}+\frac{1}{\text{0}\text{.600}\text{m}-p_1}$.

(c)

To determine

The value of $p_1$.

Answer to Problem 90AP

The value of $p_1$ is $0.30\text{m}$.

Explanation of Solution

Given info: The distance between the candle and the wall is $1.50\text{m}$. The distance by which the lens is moved is $90.0\text{cm}$. The focal length of the lens is $f$, the object distance is $p_1$.

Compare equation (2) and (4).

$\begin{array}{c}\frac{1}{p_1}+\frac{1}{1.50\text{m}-p_1}=\frac{1}{p_1+0.900\text{m}}+\frac{1}{\text{0}\text{.600}\text{m}-p_1}\\ \frac{1.50\text{m}-p_1+p_2}{p_1\left(1.50\text{m}-p_1\right)}=\frac{0.600\text{m}-p_1+p_1+0.900\text{m}}{\left(p_1+0.900\text{m}\right)\left(\text{0}\text{.600}\text{m}-p_1\right)}\\ p_1\left(1.50\text{m}-p_1\right)=\left(p_1+0.900\text{m}\right)\left(\text{0}\text{.600}\text{m}-p_1\right)\\ p_1=0.30\text{m}\end{array}$

Conclusion:

Therefore, the value of $p_1$ is $0.30\text{m}$.

(d)

To determine

The focal length of the lens.

Answer to Problem 90AP

The focal length of the lens is $0.240\text{m}$.

Explanation of Solution

Given info: The distance between the candle and the wall is $1.50\text{m}$. The distance by which the lens is moved is $90.0\text{cm}$. The focal length of the lens is $f$, the object distance is $p_1$.

The equation relating the focal length $f$ of the lens to the object distance $p_1$ is,

$\frac{1}{f}=\frac{1}{p_1}+\frac{1}{1.50\text{m}-p_1}$

Substitute $0.30\text{m}$ for $p_1$ in above equation.

$\begin{array}{c}\frac{1}{f}=\frac{1}{0.30\text{m}}+\frac{1}{1.50\text{m}-0.30\text{m}}\\ f=0.240\text{m}\end{array}$

Thus, the focal length of the lens is $0.240\text{m}$.

Conclusion:

Therefore, the focal length of the lens is $0.240\text{m}$.