Refraction through a Lens — ICSE Class 10 Physics Important Questions

Answer: (b) $0.40\,\text{m}$0.40\,m.

$P=\dfrac{1}{f}\Rightarrow f=\dfrac{1}{P}=\dfrac{1}{2.5}=0.40\,\text{m}$P=1/f f=1/P=1/2.5=0.40\,m ($=40\,\text{cm}$=40\,cm).

Answer: (c) virtual, erect and diminished.

A concave (diverging) lens forms a virtual, erect, diminished image on the same side as the object for every real object position.

Answer: (c) $2F$2F on the other side, same size.

For an object at $2F$2F ($u=-20\,\text{cm}$u=-20\,cm): $\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}-\dfrac{1}{20}=\dfrac{1}{20}$1/v=1/f+1/u=1/10-1/20=1/20, so $v=+20\,\text{cm}$v=+20\,cm. The image is real, inverted and the same size.

Answer: (c) $15\,\text{cm}$15\,cm.

Real, inverted image: $m=-3=\dfrac{v}{u}$m=-3=v/u. With $u=-20\,\text{cm}$u=-20\,cm, $v=-3u=60\,\text{cm}$v=-3u=60\,cm. Then $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{60}-\dfrac{1}{-20}=\dfrac{1}{60}+\dfrac{3}{60}=\dfrac{4}{60}$1/f=1/v-1/u=1/60-1/-20=1/60+3/60=4/60, so $f=15\,\text{cm}$f=15\,cm.

Answer: (a) For an object between $F$F and the optical centre, the emergent rays diverge and their backward projections meet on the object side, giving a virtual, erect, magnified image (as in a magnifying glass). R correctly explains A.

(i) Shape: A convex lens is thicker at the centre than at the edges; a concave lens is thinner at the centre than at the edges.

(ii) Action on light: A convex lens converges a parallel beam to a real focus (converging lens). A concave lens diverges a parallel beam so that the rays appear to come from a virtual focus (diverging lens).

(i) Nature: The power is negative, so it is a concave (diverging) lens.

(ii) Focal length: $f=\dfrac{1}{P}=\dfrac{1}{-4}=-0.25\,\text{m}=-25\,\text{cm}$f=1/P=1/-4=-0.25\,m=-25\,cm. The negative sign confirms a concave lens.

Given $u=-30\,\text{cm}$u=-30\,cm, $f=+15\,\text{cm}$f=+15\,cm, $h_o=4\,\text{cm}$h_o=4\,cm.

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{-30}=\dfrac{2-1}{30}=\dfrac{1}{30}$1/v=1/f+1/u=1/15+1/-30=2-1/30=1/30

So $v=+30\,\text{cm}$v=+30\,cm: the image is $30\,\text{cm}$30\,cm behind the lens, real and inverted.

$m=\dfrac{v}{u}=\dfrac{30}{-30}=-1$m=v/u=30/-30=-1, so $h_i=m\times h_o=-1\times4=-4\,\text{cm}$h_i=m× h_o=-1×4=-4\,cm.

The image is real, inverted, same size ($4\,\text{cm}$4\,cm), formed $30\,\text{cm}$30\,cm from the lens on the opposite side (object at $2F$2F).

Given $u=-10\,\text{cm}$u=-10\,cm, $f=-15\,\text{cm}$f=-15\,cm.

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{-15}+\dfrac{1}{-10}=\dfrac{-2-3}{30}=\dfrac{-5}{30}=\dfrac{-1}{6}$1/v=1/f+1/u=1/-15+1/-10=-2-3/30=-5/30=-1/6

So $v=-6\,\text{cm}$v=-6\,cm: the image is $6\,\text{cm}$6\,cm from the lens on the same side as the object.

$m=\dfrac{v}{u}=\dfrac{-6}{-10}=+0.6$m=v/u=-6/-10=+0.6.

The image is virtual, erect and diminished (0.6 times), located $6\,\text{cm}$6\,cm in front of the lens.

A real image on a screen is inverted, so $m=-2=\dfrac{v}{u}$m=-2=v/u, giving $v=-2u$v=-2u.

Lens formula: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$1/v-1/u=1/f

$\dfrac{1}{-2u}-\dfrac{1}{u}=\dfrac{1}{20}$1/-2u-1/u=1/20

$\dfrac{-1-2}{2u}=\dfrac{1}{20}\Rightarrow\dfrac{-3}{2u}=\dfrac{1}{20}$-1-2/2u=1/20-3/2u=1/20

$2u=-60\Rightarrow u=-30\,\text{cm}$2u=-60 u=-30\,cm.

Then $v=-2u=-2(-30)=60\,\text{cm}$v=-2u=-2(-30)=60\,cm.

The object is $30\,\text{cm}$30\,cm from the lens and the real image is $60\,\text{cm}$60\,cm from the lens on the other side.

(a) With the object between $F$F and $2F$2F, the image is real, inverted, magnified and formed beyond $2F$2F on the other side.

(b) $u=-12\,\text{cm}$u=-12\,cm, $f=+18\,\text{cm}$f=+18\,cm (object is inside the focus).

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{18}+\dfrac{1}{-12}=\dfrac{2-3}{36}=\dfrac{-1}{36}$1/v=1/f+1/u=1/18+1/-12=2-3/36=-1/36

$v=-36\,\text{cm}$v=-36\,cm. $m=\dfrac{v}{u}=\dfrac{-36}{-12}=+3$m=v/u=-36/-12=+3.

The image is virtual, erect and magnified ($3\times$3×), formed $36\,\text{cm}$36\,cm from the lens on the same side as the object.

(a) The power of a lens is the reciprocal of its focal length in metres, $P=\dfrac{1}{f\text{(m)}}$P=1/f(m); it measures the lens's ability to converge or diverge light. SI unit: dioptre (D), where $1\,\text{D}=1\,\text{m}^{-1}$1\,D=1\,m^-1.

(b) For lenses in contact, $P=P_1+P_2=(+5)+(-2)=+3\,\text{D}$P=P_1+P_2=(+5)+(-2)=+3\,D.

Focal length $f=\dfrac{1}{P}=\dfrac{1}{3}\,\text{m}=+33.3\,\text{cm}$f=1/P=1/3\,m=+33.3\,cm. Since $P>0$P>0, the combination behaves as a convex (converging) lens.

(c) $f=+33.3\,\text{cm}$f=+33.3\,cm, $u=-25\,\text{cm}$u=-25\,cm.

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{33.3}+\dfrac{1}{-25}=0.030-0.040=-0.010\,\text{cm}^{-1}$1/v=1/f+1/u=1/33.3+1/-25=0.030-0.040=-0.010\,cm^-1

$v=-100\,\text{cm}$v=-100\,cm. The image is virtual, erect and magnified, formed $100\,\text{cm}$100\,cm in front of the combination on the object's side (object lies within the focal length).

(i) The object must be placed between the optical centre and the focus, i.e. within the focal length ($<10\,\text{cm}$<10\,cm).

(ii) The image is virtual, erect and magnified, on the same side as the object.

(iii) The screen is at $10\,\text{cm}$10\,cm (the focal length). A distant object sends nearly parallel rays, so $u\to\infty$u and $\dfrac{1}{v}=\dfrac{1}{f}$1/v=1/f, giving $v=f=10\,\text{cm}$v=f=10\,cm; the real, inverted, diminished image forms at the focus.

(iv) With the object at $F$F ($u=-10\,\text{cm}$u=-10\,cm, $f=10\,\text{cm}$f=10\,cm): $\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{10}=0$1/v=1/10-1/10=0, so $v=\infty$v=. The image is formed at infinity (rays emerge parallel).