Ray Tracing and Image Formation with a Concave Lens

[binhnguyent9]

Homework Statement

If the focal length of the concave lens is -7.50 cm, at what distance d_o from the lens should an object be placed so that its image is formed 3.70 cm from the lens?
What is the magnification m produced by the concave lens described in above?
Where should the object be moved to have a larger magnification?

a) The object should be moved closer to the lens.
b) The object should be moved farther from the lens.
c) The object should be moved to the focal point of the lens.
d) The object should not be moved closer to the lens than the focal point.

Homework Equations

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The Attempt at a Solution

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[kuruman]

Complete tutorial solution

Part (a)
If the focal length of the concave lens is -7.50 cm, at what distance d_o from the lens should an object be placed so that its image is formed 3.70 cm from the lens?

We will eschew the use of numbers in favor of symbols. Use the thin lens equation. With concave lenses, I find that number substitution is less confusing if the absolute value of the focal length is used explicitly.
$$\frac{1}{d_o}+\frac{1}{d_i}=-\frac{1}{|f|}~\implies~d_o=-\frac{|f|d_i}{(|f|+d_i)}.$$A note about number substitution: One needs to be careful when substituting in the above equation. Clearly $|f|=7.50~$cm. Because the image of the concave lens is formed on the same side as the object (where light has no business being because lenses transmit light), it is virtual. Therefore, sustitute $d_i=-3.70~$cm in the equation. This will give a positive $d_o$ value.

Part (b)
What is the magnification m produced by the concave lens described in above?

The magnification is $$m=-\frac{d_i}{d_o}=+\frac{d_i+|f|}{|f|}.$$Since $d_i$ is negative, the magnification is positive and less than 1 so the virtual image is erect and diminished.

Part (c)
Where should the object be moved to have a larger magnification?
We need an expression for the magnification in terms of $d_o$ only. To do that we first find $d_i$ in terms of $d_o$ and the focal length $$\frac{1}{d_o}+\frac{1}{d_i}=-\frac{1}{|f|}~\implies~d_i=-\frac{|f|d_o}{(|f|+d_o)}$$Then $$m=-\frac{d_i}{d_o}=+\frac{|f|}{(|f|+d_o)}$$This says that the smaller $d_o$, the larger the magnification. The image is always diminished and reaches magnification 1 when the object is right up against the lens. Thanks to our commendable foresight to use the absolute value of the focal length, we see that there is no divide-by-zero problem in the expression.

Answer: a) The object should be moved closer to the lens.

The rest of the problem
I am not sure what ray diagrams A - D are supposed to show. Only one of them, A, is correctly drawn.