An object, that is 3.25 mm high is placed 16 cm in front of a lens. The image produced by this lens appears 6.0 cm behind the object, as the figure shows. (a) What is the focal length of this lens and what type of lens is it? (b) What is the height of the image? Is it real or virtual?
height of the image is 4.46875.virtual image and it is a convex lens.
To find the focal length and type of lens, we can use the lens formula. The lens formula is given by:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance (distance of the image from the lens), and
u is the object distance (distance of the object from the lens).
From the given information, we have:
u = -16 cm (negative sign indicates that the object is on the opposite side of the lens from the incident light)
v = -6.0 cm (negative sign indicates that the image is on the same side as the object)
Substituting these values into the lens formula, we have:
1/f = 1/(-6.0 cm) - 1/(-16 cm)
Simplifying the equation, we get:
1/f = -1/6.0 cm + 1/16 cm
Combining the fractions, we get:
1/f = (16 - 6)/(16 × 6) cm
Calculating the value, we have:
1/f = 10/96 cm
Taking the reciprocal of both sides, we find:
f = 96/10 cm = 9.6 cm
Therefore, the focal length of the lens is 9.6 cm.
To determine the type of lens, we can make use of the sign convention. In this case, since the image is formed on the same side as the object and is real, the lens is a converging or convex lens.
Now, let's calculate the height of the image. We can use the magnification formula, given by:
magnification (m) = height of the image (h') / height of the object (h)
Since the heights are given, we have:
h = 3.25 mm
h' = ?
From the lens formula, we know that:
h'/h = -v/u
Substituting the given values of v and u, we have:
h'/3.25 mm = -(-6 cm)/(-16 cm)
Simplifying the equation, we get:
h'/3.25 mm = 3/8
Multiplying both sides by 3.25 mm, we find:
h' = (3/8) × 3.25 mm
Calculating the value, we have:
h' ≈ 1.22 mm
So, the height of the image is approximately 1.22 mm.
Since the image is formed on the same side as the object and can be projected onto a screen, the image is real.
In summary:
(a) The focal length of the lens is 9.6 cm, and it is a converging or convex lens.
(b) The height of the image is approximately 1.22 mm, and it is a real image.