A convex lens forms inverted image of a real object on a fixed screen. The size of image is 9 cm . When lens is displaced 40 cm along principle axis it again forms a real image of size 4 cm on the screen. Focal length of the lens is .

Question

A convex lens forms inverted image of a real object on a fixed screen. The size of image is 9 cm . When lens is displaced 40 cm along principle axis it again forms a real image of size 4 cm on the screen. Focal length of the lens is . (A) 48 cm (B) 100 cm (C) 30 cm (D) 10 cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-k8khasimrtae.png" /> Since m1 m2=1 ⇒ ((h1/h0)) ((h2/h0))=1 h0= √h1 h2=6cm for first position of the lens m=-(3/2)=- (x + 40/x) ⇒ 3x=2x+80 ⇒ x= 80 u=-x , y=(.40+x.) (1/f)=(1/120)-(1/- 80) ⇒ f=48cm

Q. A convex lens forms inverted image of a real object on a fixed screen. The size of image is 9cm . When lens is displaced 40cm along principle axis it again forms a real image of size 4cm on the screen. Focal length of the lens is .

48 cm

100 cm

30 cm

10 cm

Since m1​m2​=1⇒(h0​h1​​)(h0​h2​​)=1 h0​=h1​h2​​=6cm for first position of the lens m=−23​=−xx+40​ ⇒3x=2x+80 ⇒x=80 u=−x , y=(40+x) f1​=1201​−−801​ ⇒f=48cm

Q. A convex lens forms inverted image of a real object on a fixed screen. The size of image is $9 c m$ . When lens is displaced $40 c m$ along principle axis it again forms a real image of size $4 c m$ on the screen. Focal length of the lens is .