Question

Sol. [Key Concept: Image from lens P ($f = 10$) is the object for lens Q ($f = 15$, $15$ cm away); apply the lens formula twice and $m = m_{1}m_{2}$.]

for $P\frac{1}{V} - \frac{1}{- 15} = \frac{1}{10}$

$\therefore v = + 30\text{ }cm = + 15\text{ }cm$ right of Q .

for $Q\frac{1}{{\text{ }V}^{'}} - \frac{1}{15} = \frac{1}{15}$

$${V^{'} = 7.5\text{ }cm }{m = m_{1}{\text{ }m}_{2} = \frac{30}{- 15} \times \frac{7.5}{15} = - 1 }$$