Question

An object of size 2 cm is placed at a distance of 40 cm from a glass lens, close to the principal axis of the lens, and the lens forms an image of size 2 cm. What is the size of the image and what type of image is formed if the lens and the object, in same arrangement, is immersed into water? The R. I. of the lens is 1.5 and of water is 1.33

• A. Real image, upright and 4 cm in size
• B. Virtual image, upright and 4 cm in size
• C. Real image, upright and 2 cm in size
• D. Virtual image, upright and 2 cm in size

Answer 1

Answer: (A)

$\begin{array}{c}\text{As the image is of the same size of the object}\\ \text{o its Rept on the centre of curvature of the}\\ \text{lens. Radius of wruature}40cm\\ \text{: Fows}=20cm\end{array}$

(1) $\frac{4}{2n}=(\frac{1.5}{1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$By[\frac{1}{f}=\left(\frac{{\mu }_1-1}{{\mu }_2}\right)(\frac{1}{R_1}-\frac{1}{R_2})$

(2) $\frac{1}{d1}=(\frac{1.5}{1.33}-1)(\frac{1}{R_1}-\frac{1}{R_2})\left[\begin{array}{c}\text{For lens kept in}\\ \text{water}\end{array}\right]$

Drivididing

(1) & (2)

$\frac{\frac{y}{20}}{\frac{1}{f}}=\left[0.5\right][\frac{1}{R_1}-\frac{1}{{\rho }_2}]$

$\frac{f}{20}=\left[\frac{0.17}{1.33}\right][\frac{1}{R_1}-\frac{1}{R_2}]$

$\frac{f}{20}$=$\frac{\left[0.5\right][1.33{]}_1}{\left[0.17\right]}[f\Rightarrow \frac{\left[0.5\right]\left[1.33\right]\left(20\right)}{\left[0.17\right]}]$

$V=$ image distance

$u=$ object distance

$f_{\text{water}}=75.0cm$

So By $[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}]\to$

$\frac{1}{v}-\frac{1}{(-40)}=(\frac{1}{75})\therefore V_{\text{incuter}}=-20cm$

as the sign is - ve , image is Real

And as its Real its in same side of object : its

upright, Mognification $=\frac{(-40)}{(-20)}=\frac{-u}{-v}=2$

object is of $4cm$ in size

Hence option $A$ is correct