A converging lens (f = 12.6 cm) is held 8.55 cm in front of a newspaper, the print size of which has a height of 2.10 mm.
(a) Find the image distance (in cm),
(b) The height (in mm) of the magnified print.
(a) Use 1/do + 1/di = 1/f
with f = 12.6 and do = 8.55
Solve for di (image distance)
(b) Image height = 2.1 mm * di/do
To determine the image distance and height of the magnified print, we can use the lens formula and magnification formula.
(a) To find the image distance:
The lens formula is given by:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance
u = object distance
Given:
f = 12.6 cm
u = -8.55 cm (negative because the object is placed in front of the lens)
Substituting these values into the formula, we can solve for v:
1/12.6 = 1/v - 1/-8.55
Simplifying the equation:
1/12.6 = (1/v) + (1/8.55)
Multiplying both sides by v * 12.6 * 8.55:
v = (12.6 * 8.55) / (8.55 + 12.6)
v ≈ 5.015 cm
Therefore, the image distance is approximately 5.015 cm.
(b) To find the height of the magnified print:
The magnification formula is given by:
magnification (m) = -v/u
Where:
m = magnification
v = image distance
u = object distance
Given:
v = 5.015 cm
u = -8.55 cm (negative because the object is placed in front of the lens)
Substituting these values into the formula, we can solve for magnification:
m = -5.015 / -8.55
m ≈ 0.586
The magnification value is positive, indicating an upright image.
To find the height of the magnified print, we can use the formula:
height of the image / height of the object = magnification
Let's solve for the height of the image:
height of the image = height of the object * magnification
Given:
height of the object = 2.10 mm
Substituting the values:
height of the image = 2.10 mm * 0.586
height of the image ≈ 1.2296 mm
Therefore, the height of the magnified print is approximately 1.2296 mm.
To find the image distance (di), we can use the lens formula:
1/f = 1/do + 1/di
where f is the focal length of the lens, do is the object distance, and di is the image distance.
In this case, the lens is a converging lens with a focal length of f = 12.6 cm. The object distance (do) is given as 8.55 cm.
Using the lens formula, we can rearrange it to solve for di:
1/di = 1/f - 1/do
Let's substitute the values:
1/di = 1/12.6 - 1/8.55
Now, we can calculate the value of 1/di:
1/di = 0.079 - 0.117
1/di = -0.038
To find di, we take the reciprocal of -0.038:
di = -1 / (-0.038)
di = 26.32 cm (rounded to two decimal places)
Therefore, the image distance is 26.32 cm.
Now, to find the height of the magnified print, we can use the magnification formula:
m = -di / do
where m is the magnification, di is the image distance, and do is the object distance.
Substituting the values:
m = -26.32 cm / 8.55 cm
m ≈ -3.08
The negative sign indicates that the image is inverted.
To find the height of the magnified print, we can use the magnification formula:
magnified height (hi) / object height (ho) = magnification (m)
Let's rearrange the equation to solve for hi:
hi = ho * m
The object height (ho) is given as 2.10 mm.
Substituting the values:
hi = 2.10 mm * -3.08
hi ≈ -6.45 mm
Therefore, the height of the magnified print is approximately -6.45 mm. Note that the negative sign indicates that the image is inverted.