According to the question;
Object distance (u) = -45cm;
Image distance (v) = 90cm;
By lens formula;
\(\frac1v-\frac1u=\frac1f\)
⇒ \(\frac1{90}-\frac1{-45}=\frac1f\)
⇒ \(\frac1{f}=\frac{1+2}{90}=\frac3{90}\)
⇒ \(\frac1{f}=\frac1{90}\)
⇒ f = 30 cm.
Since f = 30cm which is positive hence it is a convex lens of focal length 30cm.
Now;
Height of Flame h1= 2cm;
Magnification = \(\frac{h_2}{h_1}=\frac{v}u\)
Putting values of v and u
Magnification = \(\frac{h_2}{2}-\frac{90}{45}\)
⇒ \(\frac{h_2}{2}= -2;\)
⇒ h2 = -2 x 2 = -4
Height of image is 4 cm.
Negative sign means image is inverted.