question_answer

A student uses a convex lens to determine the width of a slit. For this he fixes the positions of the object and the screen and moves the lens to get a real image on the screen. The images of the slit width are found to be 2.1 cm and 0.48 cm wide respectively when the lens is moved through 15 cm. Therefore, the slit width and the focal length of the lens respectively, are.

A) 1 cm, 9.3 cm

B) 1 cm, 10.5 cm

C) 2 cm, 12.8 cm

D) 2 cm, 15.2 cm

Correct Answer: A

Solution :

\[{{h}_{0}}=\sqrt{{{h}_{1}}{{h}_{2}}}=\sqrt{4.1\times 0.48}\]
\[=\sqrt{0.7\times 3\times 0.16\times 3}=3\times 0.4\times \sqrt{0.7}\]\[=1.2\times \sqrt{0.7}\]\[=1\operatorname{cm}\]
For case 1
\[a\frac{{{h}_{i}}}{{{h}_{0}}}=\frac{2.1}{1}=2.1\operatorname{cm}=\frac{y}{x}\]
\[y=2.1x\]	? (i)

\[y-x=15\operatorname{cm}\]

\[2.1x-x=15\]

\[1.1x=15\]

\[x=\frac{150}{11}\operatorname{cm}=13.6cm\]

\[\frac{1}{2.1x}-\frac{1}{-x}=\frac{1}{f}\]\[\frac{2.1+1}{2.1x}=\frac{1}{f}\Rightarrow f=\frac{2.1x}{3.1}=\frac{21}{31}\times \frac{150}{11}=9.3\operatorname{cm}\]