A person suffering from myopia (near-sightedness) was advised towear corrective lens of power - 2-5 D. A spherical lens of samefocal length was taken in the laboratory. At what distance should astudent place an object from this lens so that it forms an image ata distance of 10 cm from the lens ?Draw a ray diagram to show the position and nature of the imageformed in the above case.​

The power of a lens is the reciprocal of its focal length. So, the focal length (f) of the lens can be calculated as:

f = 1 / Power f = 1 / -2.5 D f = -0.4 m or -40 cm

The negative sign indicates that the lens is a diverging lens, which is used to correct myopia.

The lens formula, which relates the object distance (u), the image distance (v), and the focal length (f) of the lens, is given by:

1/f = 1/v - 1/u

We can rearrange this formula to solve for u:

1/u = 1/f + 1/v 1/u = 1/-40 cm + 1/10 cm 1/u = -1/40 cm + 1/10 cm 1/u = -0.025 cm^-1 + 0.1 cm^-1 1/u = 0.075 cm^-1

So, u = 1 / 0.075 cm^-1 = 13.33 cm

Therefore, the student should place the object approximately 13.33 cm from the lens to form an image at a distance of 10 cm from the lens.

As for the ray diagram, I'm sorry I can't draw it here. But you can easily find examples online or in your textbook. The object will be placed to the left of the lens (since we're using the convention that distances to the left of the lens are negative), and the image will be formed on the same side as the object, indicating that it's a virtual image. The rays of light from the object will diverge after passing through the lens, but when traced back, they will appear to converge at the image point.