Phy que 12 repost paper 2 Share with your friends Share 0 Anshu Agrawal answered this Solution: For convex lens using sign convection of coordinate geometry, u=+20cm, f=−15cm So, 1/f=1/v−1/u⇒−1/15=1/v1−1/20 ⇒1/v1=1/20−1/15=3-460⇒v1=−60cm i.e., image is formed at a distance 60 cm to the left of lens L. Magnification, m1=v1/u1=−60/20=−3 This image is real and inverted. It is intercepted by the mirror. For concave mirror, u2=−60+30=−30cm,f2=+30cm So, 1/f=1/v+1/u gives 1/30=1/v2−1/30⇒1/V2=1/30+1/30=2/30⇒v2=15cm Magnification, m2=−v2/u2=(−15/−30)=+1/2 ∴ Net magnification, m=m1×m2=(−3)×(1/2)=−1.5 Size of image A'B'=−1.5×1.2cm=−1.8cm. Magnification of mirror is half and image of B fomred by convex lens is 0.6 cm above RS, so the length of image will be 1.5 cm below RS. Thus, B' will be 0.3 cm above RS and A' will be 1.5 cm below RS. 0 View Full Answer Khushi Jain answered this Please find this answer 0
