c.) A 2cm tall object is placed 12 cm to the left of a converging lens with foca

Question

c.) A 2cm tall object is placed 12 cm to the left of a converging lens with focal length having a magnitude of 15 cm. State the image location in relation to the lens, if the image is upright or inverted, if the image is real or virtual, and the magnification of the image.

d.) Now, a diverging lens with focal length having a magnitude of 20 cm is placed 10 cm to the right of the converging lens in part c. State the location of the final image formed by both lenses in relation to the diverging lens, if the final image is upright or inverted, if the final image is real or virtual, and the magnification of the final image.

Explanation / Answer

Let:
u be object distance,
v be image distance,
f be focal length,
h0 be object height,
m1 be magnification from first lens,
m2 be magnification from second lens,
h be final image height.

(c ) For the first lens:
by the len's maker's formula we have , 1 / 12.0 + 1 / v = 1 / (- 15.00)
1 / v = 1 / (- 15.00) - 1 / 12.0
v = - 6.67 cm. thus, image location is : 6.67 cm The virtual image from the diverging lens is formed on the same side of the object further , magnification of the lens
m1 = - (- 6.67) / 12.0
      = 0.56 thus, image is Virtual and upright and erected
(d) the image formed in c distace is equal to the object distnce for diverging lens thus, u = 6.67 cm focal length of the diversing lens f = 20 cm Let , v be the location of the image it is known by the formula for v as 1/ f = 1/ u + 1/ v or v = f u / u - f         = (20)(6.67) / (6.67-20)         =   - 10 cm location of the final image formed by both lenses in relation to the diverging lens is : - 10 cm image is real and up right Let:
u be object distance,
v be image distance,
f be focal length,
h0 be object height,
m1 be magnification from first lens,
m2 be magnification from second lens,
h be final image height.

(c ) For the first lens:
by the len's maker's formula we have , 1 / 12.0 + 1 / v = 1 / (- 15.00)
1 / v = 1 / (- 15.00) - 1 / 12.0
v = - 6.67 cm. thus, image location is : 6.67 cm The virtual image from the diverging lens is formed on the same side of the object further , magnification of the lens
m1 = - (- 6.67) / 12.0
      = 0.56 thus, image is Virtual and upright and erected
(d) the image formed in c distace is equal to the object distnce for diverging lens thus, u = 6.67 cm focal length of the diversing lens f = 20 cm Let , v be the location of the image it is known by the formula for v as 1/ f = 1/ u + 1/ v or v = f u / u - f         = (20)(6.67) / (6.67-20)         =   - 10 cm location of the final image formed by both lenses in relation to the diverging lens is : - 10 cm image is real and up right

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