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A glass lens with refractive index of 1.48 is placed in the air; as shown: Smaller and larger sides of the surfaces have radius of 3 cm and cm in absolute value; re...

Question

A glass lens with refractive index of 1.48 is placed in the air; as shown: Smaller and larger sides of the surfaces have radius of 3 cm and cm in absolute value; respectively. Choose true statement below:The focal length of the lens is 25 cm.The focal length of the lens is -25 cm;The focal length of the lens is -0.04 cm.The focal length of the lens is 0.04 cm:

A glass lens with refractive index of 1.48 is placed in the air; as shown: Smaller and larger sides of the surfaces have radius of 3 cm and cm in absolute value; respectively. Choose true statement below: The focal length of the lens is 25 cm. The focal length of the lens is -25 cm; The focal length of the lens is -0.04 cm. The focal length of the lens is 0.04 cm:



Answers

A single lens with two convex surfaces made of sapphire with index of refraction $n=1.77$ has surfaces with radii of curvature $R_{1}=27.0 \mathrm{~cm}$ and $R_{2}=-27.0 \mathrm{~cm} .$ What is the focal length of this lens in air? a) $17.5 \mathrm{~cm}$ c) $30.7 \mathrm{~cm}$ e) $54.0 \mathrm{~cm}$ b) $27.0 \mathrm{~cm}$ d) $40.8 \mathrm{~cm}$

Hi given in this problem there is a go on give or convex lens which may be shown like this. So for this convicts the surface, the radius of curvature Arvin for the concave surface, the radius of curvature are too and these radio are given us are one is equal to 6.0 centimetre and are too is equal to 10.0 centimetre. So this is for convex surface and this is for the concave surface. The object is stands in front of this lens. It's given us and we know object instance, he's always taken as negative. That is S is equal to minus 35.0 centimetre and image distance on the other side of these lens. The image is real. So this is on the other side of the lens and its distance will be taken as positive. That is positive 50 centimetre. Now, in the first part of the problem, we have to find a focal length of such uh lens. So using team lens equation, which is a relation among the object, is stands the industry stands and the focal length of the lens and it says one by S. Dash, The image distance minus one by S. The object distance is equal to one by F. Now, plugging in unknown values here, the fokker land will be given by one by S. Dash, which is 50 minus one by S, which is minus 35. So it comes out to be one by 50 plus one by 35 and then taking an L. C. Um This one by F comes out to be 15 in 2 35 and in enumerated, this is 35 plus 50. So finally this focal length here comes out to be 20.6 centimetre positive. So this is the answer for the first part of this given problem. The focal length of giving con cable convex lens. Now in the second part of the problem, we have to find the index of refraction of the material of rich. This concave convex lens is formed so to find it. We use lens makers formula which is again a relation for the focal land. But in terms of index of refraction and radius of curvature of both the sides and it says one by F is equal to n minus one. Record one by our one minus one by are too. So the focal length we know this is one by 20.6 is equal to end, which is missing and minus one for our one disease six centimeter for our two this is 10 centimetre. So yet it comes out to be one by 21. 6 is equal to and minus one and taking an L. C. M. Party, it comes out to be five minus tree, so this is to buy 30 which will remain 15 in the denominator. This is n minus one by 15. Hence and minus one will be given by 15 by 20.6. It will come out to be 0.73 so finally, the index of refraction of the lens material is 1.73 which is the answer for the the second part of the same problem. Thank you.

Our question says we have a thin plastic lens, has an index of refraction in of 1.67 Um, and a radio of curvature given by our one is equal to minus 12.0 centimeters and our two equals 40 centimeters. And for part A, it wants us to determine the focal length of the lens. Okay, So in order to do that, um, that is in order to find out the focal length of the lens, we're going to use the equation that says one over the focal length F is equal to the index of refraction in minus one, multiplied by the ratio of one over R one, which is negative 12 centimeters minus one over R two, which is 40 centimeters. So we need to solve for f here. So what we have to do is simply, uh, put that into the denominator. So now F is equal to one over in minus one, multiplied by one over R one minus one over R two. Okay, so now we just have to plug those values for in our one and our two under this expression, and we find that the focal length is equal to negative 13 0.8. And then the unity, of course, are centimeters that could be boxing. Is their solution for part a moving on here for Part B? Uh asked whether or not the lenses converging or diverging? Okay, So since the Focal Inc has a negative sign, um so the neck a negative focal length So we'll say a minus. Focal Inc means that the this equals a diverging linds. So negative focal length means diverging. So just box and diverging as the answer to Part B. But I wanted to give a little explanation as to why its diverging, and it's because that focal length is negative. Okay, so for part c, uh, for for part, C, D and E, it wants us to find image distances for the following object distances. So for part C, it says the object distance or we're gonna call P here is equal to infinity. Okay, well, the equation that relates image distant object distance to focal length is one over a p, so object distance plus one over cube or in his distance is equal to one over the focal length. But if P is infinity than one over P a zero. So this means that one over cube is equal to one over S. Or, in other words, Q or the image. Distance is equal to F just the focal length, which is equal to minus 13.8 centimeters. When that could be boxing is their solution for part c. Okay, Part D, we're gonna do the same thing, but this time it wants us to consider a different object distance. This time, it says that the object distance P is equal to five centimeters. Okay, so if we go back to this previous page, we have this equation which I'm gonna use a Red Star to indicate here. This one over p plus one of her cue is equal to one over f. So, using that equation we solved for Q. So cue is equal to, um, the focal length times the object distance divided by the object distance minus the focal length. So we just rearrange that equation to solve for Cuba, plugging all those values in for peas Eagle five centimeters, the focal length F. It's still negative 13.8 centimeters. Playing those values in, we find that Q is equal and negative. Three 0.67 centimeters making box set in as their solution for D. And, uh, here, the minus sign indicates that the image is formed to the left of the lens. Okay? And then for party again, the exact same thing. But now P is not equal to five cent over five centimeters. We're gonna say P is equal to 50 centimeters, but same equation. So Q is equal to s times p over T minus f clinging those values in the new value for P and the original value for F, it's still the same lens. So it still has the same focal length we find. This is equal to negative 10 0.8 centimeters. Again, the negative sign indicates that the image is formed on the left side of the lens. Then we can box set in is the solution to our question.

Late enemy is the index of reflection of surrounded medium in this problem. So the formula can be determined one by F. Is equal to end by an m minus one multiplication one by our one plus one by our two. So for a year I can buy it. One by F. Air is equal to 1.6 by one minus one multiplication one by 30 plus one by 14. On solving it further I can write the value of F. Here, F air is equal to 1200 by six multiplication seven centimetres, which is equal to 29 centimeters. Now when water is this surrounded medium I can write one by F. Air is equal to 1.6 by 1.33 minus one. My application one by 30 plus one by 14 on solving it further. Finally I can write the value of F Air is equal to 1200 by 0.203 multiplication 70 centimeters. So on solving it further I get the final answer at 84 centimeter.

As we all know that one by F is equal to and lands minus and medium by And medium multiplication and one by our 1 -1 by art. This is the fund we will use here. So simplifying it further, I can write the value of one by f is equal to 1.77 -1.33 by 1.33. Multiple choice and one by 27 centimetre minus one by minus 27 centimetres. We turn simplification. I can devalue off. One by F is equal to 0.66 by 27 centimetre. I'm solving it further. I get the value of F. He had, the centimeter is in a poolside, so the value of F is equal to 27 centimetre by 0.66 Which is equal to 40.8 cm. So option the is correct and said.


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