Answer
See the detailed answer below.
Work Step by Step
a)
For the first lens, we know that
$$\dfrac{1}{f_1}=\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}}$$
$i$ for image and $o$ for object.
When two lenses are placed in contact, the negative image of the first lens represents the object of the second lens.
$$ \dfrac{1}{d_{i1}}=\dfrac{1}{f_1}-\dfrac{1}{d_{o1}}$$
Thus,
$$ \dfrac{1}{d_{i1}}=\dfrac{1}{f_1}-\dfrac{1}{d_{o1}}=\dfrac{-1}{d_{o2}} $$
And hence,
$$ \dfrac{ 1}{d_{o2}} =-\left[\dfrac{1}{f_1}-\dfrac{1}{d_{o1}}\right]\tag 1$$
For the second lens, we know that
$$\dfrac{1}{f_2}=\dfrac{1}{d_{o2}}+\dfrac{1}{d_{i2}}$$
Plugging from (1);
$$\dfrac{1}{f_2}=-\left[\dfrac{1}{f_1}-\dfrac{1}{d_{o1}}\right]+\dfrac{1}{d_{i2}}$$
$$\dfrac{1}{f_2}+\dfrac{1}{f_1}=\overbrace{\dfrac{1}{d_{o1}} +\dfrac{1}{d_{i2}}}^{\dfrac{1}{f_T}}$$
Therefore,
$$\boxed{\dfrac{1}{f_T}=\dfrac{1}{f_1}+\dfrac{1}{f_2}}$$
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b) We know that the power of the lens is given by
$$P=\dfrac{1}{f}$$
Thus, from the boxed formula above,
$$\boxed{P_T=P_1+P_2}$$
