Questions: Magnification Optics A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm . What is the image height? (Mind your minus signs.) (Unit = cm) Submit

Solution Steps

The lens formula is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Plugging in the given values: \[ \frac{1}{16.0} = \frac{1}{24.8} + \frac{1}{d_i} \]

Rearrange the equation to solve for \( \frac{1}{d_i} \): \[ \frac{1}{d_i} = \frac{1}{16.0} - \frac{1}{24.8} \] Calculate the right-hand side: \[ \frac{1}{16.0} \approx 0.0625 \quad \text{and} \quad \frac{1}{24.8} \approx 0.0403 \] \[ \frac{1}{d_i} = 0.0625 - 0.0403 = 0.0222 \] Thus, \[ d_i = \frac{1}{0.0222} \approx 45.05 \, \text{cm} \]

The magnification \( m \) is given by: \[ m = -\frac{d_i}{d_o} \] Substitute the values: \[ m = -\frac{45.05}{24.8} \approx -1.8165 \]

The image height \( h_i \) is given by: \[ h_i = m \times h_o \] where \( h_o \) is the object height. Substituting the values: \[ h_i = -1.8165 \times 3.50 \approx -6.358 \]

Final Answer