An object is 21 cm in front of a diverging lens that has a focal length of -10 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 1.5?
Got the answer,
36.5 cm
To find the distance in front of the lens, let's use the lens formula:
1/f = 1/v - 1/u
where:
- f is the focal length of the lens,
- v is the image distance from the lens, and
- u is the object distance from the lens.
Given that the focal length f of the diverging lens is -10 cm (negative because it's a diverging lens), let's determine the initial object distance u and image distance v.
Since the object is 21 cm in front of the lens, the initial object distance (u) is +21 cm (positive since it's in front of the lens).
Now we can solve for the image distance v using the lens formula:
1/(-10) = 1/v - 1/(+21)
Simplifying the equation:
-1/10 = 1/v - 1/21
Multiply through by 210v to eliminate the denominators:
-21v = 210 - 10v
Combine like terms:
-21v + 10v = 210
-11v = 210
Divide by -11 to solve for v:
v = -210/11
v ≈ -19.09 cm
Since the image distance is negative, it means the image is formed on the same side as the object (i.e., it is a virtual image). Now that we know the initial image distance, let's find the new object distance that reduces the size of the image by a factor of 1.5.
To do this, we use the magnification equation:
magnification = -v/u
The initial magnification (magnification_initial) can be found as:
magnification_initial = -(-19.09) / 21
magnification_initial ≈ 0.91
The magnification factor is given as 1.5, so the new magnification (magnification_new) is:
magnification_new = magnification_initial / 1.5
Solving for the new object distance (u_new) using the new magnification:
magnification_new = -v/u_new
Substituting the values:
0.91 / 1.5 = -(-19.09) / u_new
Simplifying the equation:
0.6 = 19.09 / u_new
Rearranging the equation:
u_new = 19.09 / 0.6
u_new ≈ 31.82 cm
Therefore, the object should be placed approximately 31.82 cm in front of the diverging lens to reduce the size of its image by a factor of 1.5.
To find the distance in front of the lens where the object should be placed, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length
v = distance of the image from the lens (negative if the image is virtual)
u = distance of the object from the lens (positive if the object is in front of the lens)
We are given:
f = -10 cm
u = -21 cm (since the object is in front of the lens, the distance is negative)
Size of the image is reduced by a factor of 1.5, so the magnification, m = 1/1.5 = 2/3
Now, let's calculate the distance v:
1/f = 1/v - 1/u
1/-10 = 1/v - 1/-21
-1/10 = 1/v + 1/21
-1/10 = (21 + v) / (21v)
Cross-multiplying:
-21v = 10(21 + v)
-21v = 210 + 10v
-31v = 210
v = -210 / 31
v ≈ -6.77 cm (rounded to two decimal places)
Since the distance v is negative, it means that the image is virtual and located on the same side as the object (i.e., in front of the lens). To find the distance in front of the lens where the object should be placed, we take the absolute value of v:
|v| ≈ 6.77 cm
Therefore, the object should be placed approximately 6.77 cm in front of the lens so that the size of its image is reduced by a factor of 1.5.