NCERT Class 12 Physics Answers and Solutions

CHAPTER 9
9.1 v = –54 cm. The image is real, inverted and magnified. The size of the image is 5.0 cm. As u ® f, v ® ¥; for u < f, image is virtual

9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u ® ¥; v ® f (but never beyond) while m ® 0.

9.3 1.33; 1.7 cm

9.4 nga = 1.51; n wa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., r ~ 38°.

9.5 r = 0.8 × tan ic and sin 1/1.33 0.75 c i = @ , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2

9.6 n @ 1.53 and Dm for prism in water @ 10°

9.7 R = 22 cm

9.8 Here the object is virtual and the image is real. u = +12 cm (object on right; virtual)
(a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side.
(b) f = –16 cm. Image is real and at 48 cm from the lens on its right side.

9.9 v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u ® ¥, v ® f (but never beyond f while m ® 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think).

9.10 A diverging lens of focal length 60 cm

9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm = 10 cm,
fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm.
Magnifying power = 13.5.

9.12 Angular magnification of the eye-piece for image at 25 cm = 25/2.5 =1=11;|ue| =25/11cm=2.27cm vO=7.2cm
Separation = 9.47 cm; Magnifying power = 88

Please refer to attached file for NCERT Class 12 Physics Answers and Solutions