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An object is embedded in glass as shown in the following figure. If the glass has a concave face, and is embedded in water, where will the image be located? Will the image be real or virtual?
A
-1.68 cm, virtual
B
-1.68 cm, real
C
-1.83 cm, virtual
D
-1.83 cm, real
E
No image is formed
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1
Identify the refractive indices: The refractive index of the glass (n1) is 1.52, and the refractive index of water (n2) is 1.33.
Use the lens maker's formula for a single refracting surface: \( \frac{n_2}{s'} - \frac{n_1}{s} = \frac{n_2 - n_1}{R} \), where s is the object distance, s' is the image distance, and R is the radius of curvature.
Substitute the given values into the formula: \( s = 2 \text{ cm} \), \( R = 6 \text{ cm} \), \( n_1 = 1.52 \), and \( n_2 = 1.33 \).
Rearrange the formula to solve for the image distance \( s' \): \( s' = \frac{n_2 \cdot s \cdot R}{n_1 \cdot R - (n_2 - n_1) \cdot s} \).
Determine the nature of the image: If \( s' \) is negative, the image is virtual and located on the same side as the object. If \( s' \) is positive, the image is real and located on the opposite side.