In Figure $\mathrm{P} 26.38,$ a thin converging lens of focal length $14.0 \mathrm{cm}$ forms an image of the square abcd, which is $h_{c}=h_{b}=10.0 \mathrm{cm}$ high and lies between distances of $p_{d}=20.0 \mathrm{cm}$ and $p_{a}=30.0 \mathrm{cm}$ from the lens. Let $a^{\prime}, b^{\prime}, c^{\prime},$ and $d^{\prime}$ represent the respective corners of the image. Let $q_{a}$ represent the image distance for points $a^{\prime}$ and $b^{\prime}, q_{d}$ represent the image distance for points $c^{\prime}$ and $d^{\prime}, h_{b}^{\prime}$ represent the distance from point $b^{\prime}$ to the axis, and $h_{c}^{\prime}$ represent the height of $c^{\prime} \cdot(\mathrm{a})$ Find $q_{a}, q_{d}, h_{b}^{\prime},$ and $h_{c}^{\prime} .$ (b) Make a sketch of the image. (c) The area of the object is $100 \mathrm{cm}^{2}$. By carrying out the following steps, you will evaluate the area of the image. Let $q$ represent the image distance of any point between $a^{\prime}$ and $d^{\prime},$ for which the object distance is $p .$ Let $h^{\prime}$ represent the distance from the axis to the point at the edge of the image between $b^{\prime}$ and $c^{\prime}$ at image distance $q$. Demonstrate that $$\left|h^{\prime}\right|=10.0 q\left(\frac{1}{14.0}-\frac{1}{q}\right)$$ where $h^{\prime}$ and $q$ are in centimeters. (d) Explain why the geometric area of the image is given by $$\int_{q_{a}}^{q_{a}}\left|h^{\prime}\right| d q$$ (e) Carry out the integration to find the area of the image.