Question

a mirror having focal length 1m is placed below the block A. The distance between the block A and its image is 'x' m. After what time from start would the distance between the block and its image be once again 'x' m. Assume that there is no friction between the block B and the table. Mass of block B is twice that of block A.​

Answer 1

Answer:

the distance between object and its image will be again at distance "x" after t = 0.95 s

Explanation:

As we know that the object distance is u = 3 m

focal length of the mirror is f = 1 m

now we have

$\frac{1}{3} + \frac{1}{v} = \frac{1}{1}$

$\frac{1}{v} = 1 - \frac{1}{3}$

$v = \frac{3}{2} m$

so final image is at distance v = 1.5 m from the mirror

so the distance between object and its image is

$d = 3 - 1.5 = 1.5 m$

now the distance of object and image will be same when object is at u = 3/2 m and its image will form at v = 3 m

so again we can say that the block A will slide down by distance

$y = 3 - \frac{3}{2} = \frac{3}{2}$

now we know that the acceleration of A is given as

$a = \frac{mg}{m + 2m}$

$a = \frac{g}{3}$

so we will have

$\frac{1}{2}(\frac{g}{3}) t^2 = \frac{3}{2}$

$t = 0.95 s$

So the distance between object and its image will be again at distance "x" after t = 0.95 s

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Topic : Ray optics

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Answer 2

Distance between the object and its image will be again at distance "x" after t = 0.95 s

Explanation:

The object distance is u = 3 m

The focal length of the mirror is f = 1 m

We have,

$\frac{1}{3}+\frac{1}{v} =\frac{1}{1} \\\ \\ \frac{1}{v}=1-\frac{1}{3}\ \\\ v=\frac{3}{2} m$

The final image is at distance v = 1.5 m from the mirror

Therefore, the distance between the object and its image is

$d=3-1.5=1.5 m$

The distance of object and image will be the same when the object is at u = 3/2 m and its image will form at v = 3 m

Therefore, we can say that the block A will slide down by distance

$y=3-\frac{3}{2} =\frac{3}{2}$

So, the acceleration of A is given as

$a=\frac{mg}{m+2m} \\a=\frac{g}{3}$

Now we have,

$\frac{1}{2}(g/3)t^{2}=\frac{3}{2} \\t=0.95 s$

So the distance between the object and its image will be again at distance "x" after t = 0.95 s

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