QQuestionMathematics
QuestionMathematics
An object is placed 19.0 cm to the left of a converging lens that has a focal length f_1 = + 10.2 cm. How far is the image from the lens? (Give your answer to at least one decimal place.)
A screen is placed 41.0 cm to the right of the converging lens. A diverging lens is placed between the converging lens and the screen. When the diverging lens is 11.8 cm to the right of the converging lens the image formed by the lens combination is located at the screen. What is the focal length f_2 of the diverging lens?
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Answer
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Step 1:: Use the thin lens equation to find the image distance (d_i) for the converging lens.
d_i = \frac{f_1 \times d_o}{d_o - f_1}
The thin lens equation is given by:
Step 2:: Plug in the given values and calculate the image distance for the converging lens.
d_i = \frac{(+10.2 \ \text{cm}) \times (19.0 \ \text{cm})}{19.0 \ \text{cm} - 10.2 \ \text{cm}} = \boxed{30.2 \ \text{cm}}
Step 3:: Use the thin lens equation to find the image distance (d_i') for the diverging lens.
d_o' = 11.8 \ \text{cm}
The thin lens equation for the diverging lens is given by: Since the image formed by the converging lens is located at the screen, the object distance for the diverging lens is equal to the distance between the two lenses:
Step 4:: Rearrange the thin lens equation for the diverging lens to solve for the focal length (f_2):
f_2 = \frac{d_i' \times d_o'}{d_i' + d_o'}
Step 5:: Plug in the given values and calculate the focal length of the diverging lens.
f_2 = \frac{(29.2 \ \text{cm}) \times (11.8 \ \text{cm})}{29.2 \ \text{cm} + 11.8 \ \text{cm}} = \boxed{-7.58 \ \text{cm}}
Since the image formed by the lens combination is located at the screen, the image distance for the diverging lens is equal to the distance between the two lenses and the screen: Now we can plug in the values and calculate the focal length of the diverging lens:
Final Answer
The focal length of the diverging lens is f_2 = - 7.58 \ \text{cm}.
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