JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Geometrical Optics

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An object 'O' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10 s will be at ____ cm.

Answer: 400

Solution

The focal length is

f = R /2 = - 100

cm. After 10 s, the object moves

2 \times 10 = 20

cm closer, so

u = - (100 - 20) = - 80

cm. Using the mirror equation

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} + \frac{1}{- 80} = \frac{1}{- 100}

. Solving:

\frac{1}{v} = \frac{1}{- 100} + \frac{1}{80} = \frac{- 4 + 5}{400} = \frac{1}{400}

. Therefore

v = 400

cm (real image behind the mirror).

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer

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