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This is second of our many quizzes on Mirrors. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Mirrors Quiz 2

physics

A convex mirror is placed at the bottom of a bucket of depth 30cm. The bucket is equally filled with liquid 1 and liquid 2, an object is placed at the top of the bucket, find the apparent distance of the object. $[$refractive index of liquid one ${{\mu }_{1}}=\dfrac{3}{2}$ and for two ${{\mu }_{2}}=3]$ 

 Due to two liquids present between the mirror and object, the distance of the object from the mirror will change with respect to the mirror and the new apparent distance will be the distance of the object from the mirror. 

  Apparent distance in liquid or multiple liquids is given by $ \\ d'=\dfrac{d}{2{{\mu }_{1}}}+\dfrac{d}{2{{\mu }_{2}}}\\ $ So apparent distance of object from a convex mirror will be $ \\ d'=\dfrac{d}{2{{\mu }_{1}}}+\dfrac{d}{2{{\mu }_{2}}}\\\therefore d'=\dfrac{30}{2\times \dfrac{3}{2}}+\dfrac{30}{2\times 3}=15\,cm$

If a person is standing before a concave mirror between focal length and centre of curvature, but since he can’t see his image, he used a plane mirror to see his real image formed by a concave mirror. What will be the location of the plane mirror that is able to see that image by person? 

 The position of the image of the person can also be calculated by the lens formula to calculate the distance of the image from the mirror. 

  The person is standing between the radius of curvature and focus,  the image will be inverted and formed behind the person and before the radius of curvature, if a plane mirror is placed in front of the person or between mirror and person’s position, the person can see his image from the image formed by a concave mirror. Therefore, the position can be between focus and person or between focus and mirror.

An object is placed before a plane mirror at a distance of $40\,cm$, after some time the object starts moving at $2\,cm/sec$ and the mirror starts moving at $5\,cm/sec$ in the right direction. Find the distance between the object and its image at t = 5 sec. 

 Objects and mirrors are moving at a certain speed so by observing carefully, we can say that the image is also moving and the speed of it can be calculated by measuring the speeds of the mirror and object. 

 The distance covered by mirror in 5 sec will be $ \\{{D}_{m}}=vt=5\times 5=25\,cm\\ $ Distance travelled by object in 5 sec will be $ \\ \Rightarrow {{D}_{o}}=vt=2\times 5=10\,cm\\ $ Initial distance between object and plain mirror was 40 cm and final distance will be $ \\d=25-10+40=55cm\\ $ Therefore,  the image will be formed at a distance of 55 cm from the plane mirror so, the total distance between object and its image will be 110 cm.

Object B is placed at distance D from a convex lens of focal length 30 cm and the image formed is virtual. If the image formed is magnified 5 times. Now the objects form a different image due to convex mirrors of focal length 25cm. The distance between mirror and lens is 49 cm. Find the distance of the final image formed due to the mirror.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/84792f5e-7c5f-4979-857e-9d604479fb494905073355928275495.png"/>

First, the mirror will form an image and by applying the lens maker formula, we can calculate the distance of the object then by the mirror formula, the required answer is calculated. 

 Lens maker formula $ \\ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\ $ Image from a convex mirror is magnified 5 times so $ \dfrac{v}{u}=5$ $ \\ $ By putting value of final velocity from above equation in lens maker formula we have $ \\ \dfrac{1}{30}=\dfrac{1}{5u}-\dfrac{1}{u}=-\dfrac{4}{5u}\\\Rightarrow u=-24cm\\ $ Now distance of object from mirror is $ \\ d=49-24=25cm\\\Rightarrow \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\\\Rightarrow \dfrac{1}{-25}=\dfrac{1}{v}+\dfrac{1}{25}\\\therefore v=-12.5\,cm$

An object is placed in front of a plane mirror at a distance of $50\,cm$. Now the object starts moving with $25\,cm/sec$ towards the mirror and the mirror starts moving with $7\,cm/sec$ towards the object. If a man is sitting on a mirror the relative velocity of the object with respect to man will be-

 In the reference frame, one can find himself at rest and the others will be in relative motion with respect to him.

 Man is sitting on a plane mirror moving with a velocity of 7 m/s, so in the frame or man mirror is at rest and the object is moving with some velocity. The velocity of an object with respect to man will be given by the summation of the velocity of the object and that of the mirror is $v=25+7=32m/s$.