• Class 11 Physics Demo

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a concave mirror has a focal length of

Question:A concave mirror has a focal length of 13 cm. This mirror forms an image located 52 cm in front of the mirror. What is the magnification of the mirror? (Include the sign.)

Answers:I hope you know how to use the equation: 1/u + 1/v = 1/f u = Distance of object, +ve if real, -ve if virtual. v = Distance of image, +ve if real, -ve if virtual. f = Focal length, +ve if concave mirror, -ve if convex mirror. For object, In front of mirror = real (+ve) Behind the mirror = virtual (-ve)*** ***This is the special case for object to behind the mirror, but it's not impossible. For image, In front of mirror = upright wrt. object = virtual (-ve) Behind the mirror = inverted wrt. object = real (+ve) *wrt. = with respect to Now, come back to calculation, 1/u + 1/(-52) = 1/13 1/u = 1/13 + 1/52 1/u = 5/52 u = +10.4 cm Question wants magnification, so by definition, M = v/u M = (-13)/(+10.4) M = -1.25 Negative sign means image is inverted wrt. object and 1.25 mean image is large than object by 1.25. Hope it helps you.

Question:A concave mirror has a focal length of 11 cm. The mirror forms an image located 32 cm in front of the mirror. What is the magnification of the mirror?

Answers:1/f = 1/i + 1/o where f = focal length o = distance of object from lens i = distance of image from lens 1/o = 1/11 - 1/32 o = 16.76 cm m = -32/16.76 m = -1.91, where the negative sign indicates the image is inverted

Question:A concave mirror has a focal length of 20 cm. What is the magnification if the object and image distance are 10 cm and -20 cm respectively?

Answers:well, magnification = v/u where 'v' is the image distance 'u' is the object distance m = -20/10 = -2. where negative sign represents that it is a virtual image.

Question:A concave mirror has a focal length of 31.6 cm. Determine the object position for which the resulting image is upright and six times the size of the subject. I have gotten 36.86cm and was told that this was an incorrect answer.

Answers:since its a concave mirror, f = -31.6 cm let object position be u and image position v magnifivation m = - v/u = h'/h h is abject size n h' image size resulting image is uprigth so h' is +ve and is 6h - v/u = 6 or v = -6u by mirror formula 1/v+1/u=1/f 1/u = 1/f-1/v 1/u = 1/f + 1/6u 5/(6u) = 1/f u = f * 5/6 u = 31.6 * 5/6 u= 26.3 cm(approx) its kept at a distance 26.3 cm from the pole of the concave mirror.