Answer :
The image distance for the object placed 26 cm from the lens is -3.11 cm, and the magnitude of the magnification is 0.12.
The image distance can be calculated using the lens formula:
[tex]1/f = 1/u + 1/v[/tex]
Where f is the focal length of the lens, u is the object distance, and v is the image distance.
Given that the focal length of the diverging lens is 35.8 cm and the object distance is 26 cm, we can substitute these values into the formula:
[tex]1/35.8 = 1/26 + 1/v[/tex]
To solve for v, we need to first find the common denominator:
[tex](26v)/(26v) * (1/35.8) = (26v)/(26v) * (1/26) + (26v)/(26v) * (1/v)[/tex]
Simplifying the equation:
[tex]v/35.8 = 1/26 + 1/v[/tex]
To combine the fractions, we find the common denominator again:
[tex]v/35.8 = (v + 35.8)/26v[/tex]
Cross-multiplying:
[tex]26v^2 = 35.8(v + 35.8)[/tex]
Expanding the equation:
[tex]26v^2 = 35.8v + 35.8^2[/tex]
Rearranging the equation:
[tex]26v^2 - 35.8v - 35.8^2 = 0[/tex]
Using the quadratic formula:
[tex]v = (-b ± √(b^2 - 4ac)) / (2a)[/tex]
[tex]where a = 26, b = -35.8, and c = -35.8^2[/tex]
Plugging in the values:
[tex]v = (-(-35.8) ± √((-35.8)^2 - 4(26)(-35.8^2))) / (2(26))[/tex]
Calculating the equation:
[tex]v = (35.8 ± √(1284.64 + 37729.12)) / 52[/tex]
[tex]v = (35.8 ± √(39013.76)) / 52[/tex]
[tex]v = (35.8 ± 197.52) / 52[/tex]
Therefore, there are two possible solutions:
[tex]v1 = (35.8 + 197.52) / 52 = 233.32 / 52 = 4.48 cm[/tex]
[tex]v2 = (35.8 - 197.52) / 52 = -161.72 / 52 = -3.11 cm[/tex]
Since the object is placed on the same side as the lens, the image distance should be negative. So, the image distance is -3.11 cm.
To find the magnitude of the magnification, we can use the formula:
m = -v/u
Plugging in the values:
[tex]m = -(-3.11) / 26 = 0.12[/tex]
Therefore, the magnitude of the magnification for this lens is 0.12.
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