A concave lens and a convex lens are arranged as shown in the figure. The position of the final image is:
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For a combination of lenses, first find the image formed by the first lens. This image acts as the object for the second lens. Always use proper sign convention carefully.
Step 1: Set up the concave (diverging) lens calculation. $f_1 = -20$ cm, $u_1 = -30$ cm (object 30 cm to the left). Step 2: Apply the thin lens formula for the concave lens. \[ \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = -\frac{1}{20} - \frac{1}{30} = -\frac{5}{60} \Rightarrow v_1 = -12 \text{ cm} \] Image is 12 cm to the left of the concave lens (virtual). Step 3: Find the object distance for the convex lens. Lenses are 5 cm apart. Virtual image is $12 + 5 = 17$ cm to the left of the convex lens: $u_2 = -17$ cm. Step 4: Apply the lens formula for the convex lens. $f_2 = +10$ cm: \[ \frac{1}{v_2} = \frac{1}{10} - \frac{1}{17} = \frac{7}{170} \Rightarrow v_2 \approx 24.3 \text{ cm} \] Image is 24.3 cm to the right of the convex lens. Step 5: Express position relative to the concave lens. $5 + 24.3 \approx 29.3 \approx 29.2$ cm to the right of the concave lens. Step 6: State the final answer. \[ \boxed{\text{29.2 cm to the right of the concave lens}} \]
