Join thousands of students who trust us to help them ace their exams!
Multiple Choice
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
3 Comments
Verified step by step guidance
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.