You can get the object height from the magnification. Since the image is [itex]20[/itex] times smaller than the object, [itex]M=1/20[/itex]. We also know that [itex]H_i=2 \, m[/itex].Samr28 said:
Multiply the equation [tex]\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}[/tex] through by [itex]d_o[/itex]. This gives you the equation [tex]\frac{d_o}{f}=\frac{d_o}{d_o}+\frac{d_o}{d_i}[/tex] [itex]d_o/d_o=1[/itex] and [itex]d_o/d_i=1/M[/itex] Solve for the remaining [itex]d_o[/itex].andrevdh said:
A converging lens is a type of lens that is thicker in the middle and thinner at the edges. It is also known as a convex lens. This type of lens is used to bring parallel rays of light together to a single point, known as the focal point.
The focal length of a converging lens is the distance from the center of the lens to the focal point. It is represented by the letter 'f' and is measured in meters (m).
In the formula for a converging lens, Ho represents the object height, which is the height of the object being viewed. Hi represents the image height, which is the height of the image formed by the lens.
To solve for all variables in the formula for a converging lens, you can use the equation: 1/f = 1/Hi + 1/Ho. Plug in the values for f, Hi, and Ho and solve for the remaining variable. Remember to use the correct units for each variable.
Yes, a converging lens can have a negative focal length. A negative focal length means that the focal point is located on the opposite side of the lens from where the light is coming from. This is known as a virtual image, where the light rays do not actually converge at a point, but appear to be coming from a specific point when viewed through the lens.