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Consider the situation skteched. A fish ...

Consider the situation skteched. A fish lies 40 cm under the surface of water `(mu_(water)=4//3)`, 80 cm above a concave spherical mirro whose radius of curvature is 120cm. An observer looking down from above in the air `(mu_(air)=1)` sees two images of fish. Find distance (in meters) between these two images.

Text Solution

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The observer will observe two images, Once by direct observatino and second will be reflected image forward by mirror.
a. Observer directly observe the fish.
Apparent depth
`=(Real depth)/mu_(rel)`
`=h_(0)/((mu_(water)/mu_(air)))=40/(40//3)=30cm`
The observer will observe fish 30 cm below the water surface when he see directly.
b. Observe obser the reflected image form by mirror.
Considering reflectino from mirror,
u=-80 cm, `f=-R/2=(-120)/2=-60cm`
Using `1/v+1/u=1/f`
`rArr 1/v+1/((-80))=1/((-60))rArrv=-240cm`
i.e., 240 cm above mirror or 120 cm above water surface.
Now this image formed by mirror will acts as object for top water surface. Now rays are moving from water to air.
The position of this 'image' as seen by observer observin from top of the water surface
Apparent depth `h^(') = (h_(real))/(mu_(rel))=(h_(rel))/(mu_(water)//mu_(air))`
`=(120)/(((4//3)/1))=90 cm` above water surface
Hence, distance between two images
`d=(600+90)=150cm= 1.5m`
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