A convex lens forms an image of an object on a screen ,The height of the image is$9cm$The lens is now disiplaced between the object on the screen.the height of this image is$4cm$.the distance between the object and the screen is $90cm$a. The distance between the two positions of the lens is $30cm$ b. The distance of the object from the lens in its first position is$36cm$ c. The height of the object is $6cm$ d. The focal length of the lens is$21.6cm$

Sure, here are the step by step solutions: 1. Using the formula for the height of a trapezium, we have: $h=\sqrt{h_1{h}_2}\Rightarrow h=6\text{ cm}$ 2. Using the formula for the magnification of a lens, we have: $m=\frac{h_1}{h}=\frac{v}{u}\Rightarrow \frac{-9}{6}=\frac{D-x_1}{-x_1}$ 3. Solving for $x_1$, we get: $x_1=36\text{ cm}$ 4. Using the fact that the distances from the object to the lens and from the image to the lens add up to the focal length, we have: $x_2=D-x_1\Rightarrow x_2=54\text{ cm}$ 5. Using the formula for the distance between the object and the image, we have: $d=x_2-x_1\Rightarrow d=18\text{ cm}$ 6. Using the formula for the focal length of a lens, we have: $f=\frac{D^2-d^2}{4D}\Rightarrow f=\frac{90^2-18^2}{4\times 90}$ 7. Simplifying the expression, we get: $f=\frac{(90+18)(90-18)}{4\times 90}\Rightarrow f=\frac{108\times 72}{4\times 90}=\frac{3\times 72}{10}=21.6\text{ cm}$