\begin{tabular}{l} A diverging'lens can form a real image. \\ True \\ False \\ Question 21 \\ Whaby lionturtle sits 50 cm from a converging lens. It's image appears at -80 cm . \\ What is the lens's focal length? \\ \hline\end{tabular}

Let's break down the problem step by step. ### Known Conditions: 1. The lens is a converging lens. 2. The object distance \( d_o = 50 \, \text{cm} \) (positive because it's on the same side as the incoming light). 3. The image distance \( d_i = -80 \, \text{cm} \) (negative because it's a virtual image formed on the opposite side of the lens). ### Lens Formula: The lens formula relates the object distance, image distance, and focal length \( f \) of a lens: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] ### Step 1: Substitute the known values into the lens formula. We will substitute \( d_o = 50 \) and \( d_i = -80 \) into the lens formula. ### Step 2: Calculate the focal length \( f \). Let's perform the calculation: \[ \frac{1}{f} = \frac{1}{50} + \frac{1}{-80} \] Now, we will calculate this expression. Calculate the value by following steps: - step0: Calculate: \(\frac{1}{50}+\frac{1}{\left(-80\right)}\) - step1: Remove the parentheses: \(\frac{1}{50}+\frac{1}{-80}\) - step2: Rewrite the fraction: \(\frac{1}{50}-\frac{1}{80}\) - step3: Reduce fractions to a common denominator: \(\frac{8}{50\times 8}-\frac{5}{80\times 5}\) - step4: Multiply the numbers: \(\frac{8}{400}-\frac{5}{80\times 5}\) - step5: Multiply the numbers: \(\frac{8}{400}-\frac{5}{400}\) - step6: Transform the expression: \(\frac{8-5}{400}\) - step7: Subtract the numbers: \(\frac{3}{400}\) Calculate or simplify the expression \( 1/(3/400) \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{\frac{3}{400}}\) - step1: Multiply by the reciprocal: \(\frac{400}{3}\) ### Step 3: Calculate the focal length \( f \) From the previous calculation, we found: \[ \frac{1}{f} = \frac{3}{400} \] To find \( f \), we take the reciprocal: \[ f = \frac{400}{3} \approx 133.33 \, \text{cm} \] ### Conclusion: The focal length of the converging lens is approximately \( 133.33 \, \text{cm} \). ### Additional Note: Regarding the first statement about a diverging lens forming a real image, the answer is **False**. A diverging lens cannot form a real image; it only forms virtual images.