Open Question
A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (b) Where is the ? Is the real or virtual?
33. Geometric Optics
Mirror Equation
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A 4 cm tall object is placed in 15 cm front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?
A
si = 7.5cm; Real; Inverted, 2cm
B
si = 15cm; Real; Inverted, 4cm
C
si = 0.13 cm; Real; Inverted, 0.0087cm
D
si = -7.5cm; Real; Inverted, -2cm
Verified step by step guidance1
Identify the given values: object height (ho) = 4 cm, object distance (do) = 15 cm, and focal length (f) = 5 cm.
Use the mirror equation to find the image distance (si): \( \frac{1}{f} = \frac{1}{do} + \frac{1}{si} \). Substitute the known values into the equation.
Rearrange the mirror equation to solve for the image distance (si): \( \frac{1}{si} = \frac{1}{f} - \frac{1}{do} \). Calculate \( \frac{1}{si} \) and then take the reciprocal to find si.
Determine the nature of the image: If si is positive, the image is real and inverted. If si is negative, the image is virtual and upright.
Calculate the image height (hi) using the magnification formula: \( m = \frac{hi}{ho} = -\frac{si}{do} \). Substitute the known values to find hi and determine if the image is upright or inverted based on the sign of hi.
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