a)ta có :
D = \(\dfrac{1}{f}\) = 3 (dp) → f =\(\dfrac{1}{3}m\)= \(\dfrac{100}{3}cm\)
b)d1 = 10cm
→d1' = \(\dfrac{d_1.f}{d_1-f}\) = \(\dfrac{10.\dfrac{100}{3}}{10-\dfrac{100}{3}}\)= \(\dfrac{-100}{7}\) cm
→ảnh là ảnh ảo
K = \(\dfrac{d_1'}{d_1}\) = \(\dfrac{-100}{7.10}\) = \(\dfrac{-10}{7}\)