SOLVED:A coin is resting on the bottom of a tank of water \left(n_{\mathrm{W}}=1.33\right) 1.00 \mathrm{m} deep. On top of the water floats a layer of benzene \left(n_{b}=1.50\right) which is 20.0 \ma

Video Transcript

a mint is placed at a depth of one thousand from the absolve surface of water. About the absolve airfoil of body of water a layer of Benjamin is floating. The thickness of the layer of Benjamin is 20 centimeter or 0.2 m. The deflective index of water system is given to us. That is N. W. is adequate to 1.33. The deflective exponent of Benjamin that is denoted by N. B. That is adequate to 1.5. now we have to find on seeing from the top airfoil how army for the liberation of rwanda below the acme airfoil the coin will appear. so first if we see the mint from the belgian average, then we ‘ll find a particular apple iraqi national congress of the coin with deference to the bench in medium. then if we again see the coin from the uh above the crouch medium that is from publicize medium. then we ‘ll find that the former case after. In affected role will be the actual patient for the bench in character. And then there will be another amphetamine impatient with deference to the air medium. thus nowadays first beginning with. Okay for observation for observation from vengeance medium. For observation from bending medium we can write that the apparent acme in the first think by the actual altitude, actual stature in the first case is equal to the ratio of the deflective index of body of water medium with respect to uh flex. That is basically going to be peer to the proportion of the the refractive index of Benjamin divided by the density the deflective index of water system. Because we are seeing this from bending. So we have to take the proportion of the referral from bending by the uh deflective index of water system. indeed following this we can write that the apparent height of the mint with deference to Benji million. That is the inaugural casing that have taken is equal to the actual height. Uh The electoral death of the coin with the spirit of Benjamin medium. Into in refrigeration crouch by deflective index of body of water. No actual height of actual death of the coin with the special mention medium is uh and if the one millimeter into N. B. S. 1.5 divided by 1.33 then this is going to be adequate to age a parent in the beginning case is equal to 1.1278 m. so on seeing from the Benjamin medium we will see that the neologism nowadays of 12 1.1278 m. So it will be somewhere below the water degree of given situation with deference to the bench in medium. now then uh if we now observe from the upper berth grade of Benjamin in that case the actual altitude or actual death of the uh coin is going to be equal to. so now we ‘ll right with deference to yeah medium above bingeing surface. The actual status of the coin Is that when we do nothing by uh page actual in the second case that is going to be called the wealth of the flex medium which is equal to 0.2 molarity plus the apparent uh death are apparent height of the uh Position with the speed of bending million which is equal to 1.1278. then this is going to be equal to The actual higher. In the second lawsuit is going to be peer to 1.3278 m. now if we want to find the apparent position of the mint with esteem to the air metier then we can write apparent position. apparent placement of the coin adds seen from the rid sarah fists above vengeance is age apparent in the second case that we have to find divided by age actual. In the second case is equal to the deflective index of air by the deflective index of the benji in medium. That is basically N by N. B. now from here we can write that the apparent position of the mint as in from air out is adequate to age actual in the second casing into N. That is one By the different denote in this of the Belgian is 1.5. No they hate after. In the second base case is nothing but 1.3278. So if I put this sleep together is with a dependable page, A parent in the second case is equal to one 3278 divided by 1.5. So that will be equal to age is different. In the second case is equal to 0.8852 m. sol this is the position of the uh coin from the upper surface, so after the 0.852 meter from the upper surface will be uh seeing the mint if we see directly from from the uh top side of the substitute, depending million.

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