An object is placed 3cm in front of a diverging lens of focal length of 6cm. the heihgt of the object is 1.5 cm. calculate the image distance and height
-1/6-1/3=-1/2
image distance=-2
hi/ho=-di/do
hi/1.5=-(-2)/3
hi=(2/3)x(3/2)
hi=1
i want to know the formula according to conventions as i saw two different systems as i studied.
For convex and concave lens and for concave and convex mirror
the formula is
1/ di + 1/do = 1/f
m = hi/ho = - di /do
Note that different conventions are in practice .
Accordingly the formula will change .
For the above formula the following sign conventions are used.
• f is + if the lens is a double convex lens (converging lens)
• f is - if the lens is a double concave lens (diverging lens)
• di is + if the image is a real image and located on the opposite side of the lens.
• di is - if the image is a virtual image and located on the object's side of the lens.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is - if the image an inverted image (and therefore, also real)
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For concave and convex mirror
• f is + if the mirror is a concave mirror
• f is - if the mirror is a convex mirror
• di is + if the image is a real image and located on the object's side of the mirror.
• di is - if the image is a virtual image and located behind the mirror.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is - if the image an inverted image (and therefore, also real)
To calculate the image distance and height in this scenario, we can use the lens formula and the magnification formula.
The lens formula is given by:
1/f = 1/v - 1/u
Where:
- f represents the focal length of the lens,
- v represents the image distance,
- u represents the object distance.
In this case, the focal length (f) is given as 6 cm, and the object distance (u) is given as -3 cm (-3 cm because the object is placed in front of a diverging lens).
Let's substitute these values into the lens formula:
1/6 = 1/v - 1/-3
To simplify the equation further, we can get rid of the fractions:
1/6 = (1/v) + 1/3
Now, let's find the common denominator by multiplying both sides of the equation by 6v:
v = 6 + 2v/3
Multiply both sides by 3:
3v = 18 + 2v
Subtract 2v from both sides:
v = 18 cm
So, the image distance (v) is 18 cm.
Now, let's calculate the image height using the magnification formula:
magnification (m) = -v/u
Where:
- v represents the image distance,
- u represents the object distance.
Substituting the values:
m = -18 / -3
m = 6
The magnification (m) in this case is 6, which means the image is magnified 6 times compared to the object.
Now, let's calculate the image height (h') using the formula:
h' = m * h
Where:
- h' represents the image height,
- m represents the magnification,
- h represents the object height.
Substituting the values:
h' = 6 * 1.5 cm
h' = 9 cm
So, the image height (h') is 9 cm.