Question

A collector is using a magnifying lens with a radius of curvature of 21.0 cm to examine the finer details of a butterfly with a magnification of 5.10. At what distance between the butterfly and the lens does she achieve the desired magnification? O 13.1 cm 0 12.6 cm 8.40 cm 8.80 cm 16.8 cm

Answer 1

(1) Given: lens > Radius of curvature of lens (R)= 21.0cm magnifying Conven Lens R So, focal length (f) = 21 2 2 = +10.5 cm (conven) 나 Magnification (m) = 5.10 We know that, m = u $13813 5.1 = رأ۔ v = 5.1u find Butterfly distance (u) = ? +-+- 11 f 10.5 5.14

los = á (5-0) -4.1 1] 10.5 5.14 4.1 501 u = -10.5 u -0.40 cm ANSWER (2) Given- Cancave mirror Radius of curvature (R) = loocm R So, focal length (f) = R 100 2 2 f = -50 cm (concave) Distance of image (V) = -2.7 m =-27ocm ( Inverted) Find, object distance (4) = ? We know that 5 te tú st -50) (-270) u - - 50 270 27 -5 - u) 1350 1350 22 u u=-61.4cm Option @ is correct front of mirror

In both of the question "u" is coming with negative sign it's mean that object is placed in front of lens or mirror.

For that we use the sign convention-

When light ray traveling direction is opposite to the distance measurement direction then the distance will be taken as "negative".