a compound microscope has an eyepiece of focal length 10 c.m and an objective of focal length 4c.m.the magnification of an object is placed at a distance of 5 c.m from the objective so that the final image is formed at the distance of distinct vision (20 c.m) is
To calculate the magnification of a compound microscope, we can use the formula:
Magnification = (-ve focal length of objective) / (+ve focal length of eyepiece)
From the given information, the focal length of the eyepiece (Ey) is 10 cm, and the focal length of the objective (Ob) is 4 cm. Plugging these values into the formula, we get:
Magnification = (-4 cm) / (+10 cm)
Magnification = -0.4
Since the magnification is negative, it indicates an inverted image.
Now, to determine the position of the object (O) that gives a final image at the distance of distinct vision (20 cm), we can use the formula:
1/f = 1/v - 1/u
Where:
f = focal length of the objective
v = final image distance (negative if inverted)
u = object distance
By rearranging the formula, we have:
1/u = 1/f - 1/v
Plugging in the values:
f = 4 cm
v = -20 cm (negative indicating inverted image)
1/u = 1/4 cm - 1/-20 cm
Simplifying further, we get:
1/u = 5/20 - 1/4
1/u = 1/4 - 1/4
1/u = 0
This implies that the object distance (u) is at infinity. It means the object is placed at the focal point of the objective lens.
Thus, the object should be placed at a distance of infinity from the objective lens to obtain a final image at the distance of distinct vision (20 cm) with a magnification of -0.4.