Question

A telescope has an objective lens of focal length 150 cm and an eyepiece with focal length 1.5 cm. If this telescope is used to see a 50 m tall building at a distance of 1.5 km, what is the height of the image of the building formed by the objective lens?

• A. 10 cm
• B. 5cm
• C. 20 cm
• D. 15 cm

Answer 1

Answer: (B)

For objective (convex) lens $ \frac{2{{\eta }^{2}}}{g{{a}^{2}}} $ Here, $ \frac{{{\eta }^{2}}}{g{{a}^{2}}} $ $ \frac{{{\eta }^{2}}}{2g{{a}^{2}}} $ $ \frac{{{\eta }^{2}}}{4g{{a}^{2}}} $ $ {{t}_{1}} $ $ {{t}_{2}} $ $ 2g({{t}_{1}}+{{t}_{2}}) $ $ \frac{g}{4}{{({{t}_{1}}+{{t}_{2}})}^{2}} $ $ g/4({{t}_{1}}{{t}_{2}}) $ Magnification $ 2g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{4} \right)}^{2}} $ $ R/4 $ $ R/2 $ $ R/8 $ $ \left\{ \frac{{{(1+e)}^{2}}}{1+{{e}^{2}}} \right\}\sqrt{\frac{gh}{2}} $