Sapling Learning Map Determine the image distance for an object 7.000 cm from a

Question

Sapling Learning Map Determine the image distance for an object 7.000 cm from a diverging lens of radius of curvature 5.600 cm and index of refraction 1.800. (Express your answer as a positive quantity.) Number 12.7 cm R- 5.600 c object What is the magnification of the object? (Be sure to insert the proper sign if needed.) Number 3.76 What is the nature of the image? O Inverted and smaller O Upright and smaller There is additional feedback er available View this feedback by clicking on the bottom divider bar on the divider Incorrect bar again to hide the additional feedback

Explanation / Answer

Image position, made by lens is given by 1/v - 1/u = 1/f = (n-1)(1/r1 - 1/r2 )

Here v is image coordinate, u is object coordinate, n is refective index of lens material, r1 and r2 are coordinates of center of curvature of two curved sufaces. origin of coordinates is taken at center of lens, and +ve direction towards right. As per problem u = -7 n = 1.8 r1 = -5.6 and r2 = + 5.6

So 1/v + 1/7 = 0.8 (- 2/5.6)

hence v = - 2.33 cm hence distance of image is 2.33 cm

Magnification is v/u here again v and u are image and object coordinates

hence magnification is -2.33/(- 7) = + 1/3

As magnification is +ve image is upright and as it is less than one, image is small.

2) A -ve focal length or diverging lens always mkes virtual image of real object. Hence image of ring will be virtual.

I/O = v/u = f/(f+u), we get second equality from lens equation 1/v-1/u =/1/f. Here I is image size and O is object size.

hence I/3.23 = -5.35/(-5.35 - 12.7) hence I = 0.96 mm

As magnification = v/u = f/(f+u) is +ve, image is upright

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.