To solve this problem, we can use the lens formula which is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

Given that the image is 10 times the size of the object, the object distance (u) and image distance (v) are related by the magnification formula, which is -v/u = magnification. In this case, the magnification is -10 (negative because the image is inverted).

We can rearrange the magnification formula to find u = -v/10.

We know that the screen (where the image is formed) is 8.0 m from the lens, so v = -8.0 m (negative because the image is on the same side of the lens as the light source).

Substituting v = -8.0 m into the equation for u gives u = -(-8.0 m)/10 = 0.8 m.

Now we can substitute u = 0.8 m and v = -8.0 m into the lens formula to find the focal length:

1/f = 1/(-8.0 m) - 1/(0.8 m) = -0.125 m⁻¹ + 1.25 m⁻¹ = 1.125 m⁻¹.

Therefore, the required focal length of the lens is f = 1/(1.125 m⁻¹) = 0.89 m.

Question

To solve this problem, we can use the lens formula which is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

Given that the image is 10 times the size of the object, the object distance (u) and image distance (v) are related by the magnification formula, which is -v/u = magnification. In this case, the magnification is -10 (negative because the image is inverted).

We can rearrange the magnification formula to find u = -v/10.

We know that the screen (where the image is formed) is 8.0 m from the lens, so v = -8.0 m (negative because the image is on the same side of the lens as the light source).

Substituting v = -8.0 m into the equation for u gives u = -(-8.0 m)/10 = 0.8 m.

Now we can substitute u = 0.8 m and v = -8.0 m into the lens formula to find the focal length:

1/f = 1/(-8.0 m) - 1/(0.8 m) = -0.125 m⁻¹ + 1.25 m⁻¹ = 1.125 m⁻¹.

Therefore, the required focal length of the lens is f = 1/(1.125 m⁻¹) = 0.89 m.

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