Q.41. An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another con

Q: What is the correct answer for JEE Main Physics Q41?

The correct answer is B. For lens P (convex, $f_P = +10$ cm): Object distance $u_P = -15$ cm (left of P). Using lens formula $\frac{1}{v_P} - \frac{1}{u_P} = \frac{1}{f_P}$: $\frac{1}{v_P} - \frac{1}{-15} = \frac{1}{10} \Rightarrow \frac{1}{v_P} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$. So, $v_P = +30$ cm. The image formed by lens P is real, 30 cm to the right of P. Now, for lens Q (convex, $f_Q = +

Q: Q.41. An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final…

The correct answer is B. For lens P (convex, $f_P = +10$ cm): Object distance $u_P = -15$ cm (left of P). Using lens formula $\frac{1}{v_P} - \frac{1}{u_P} = \frac{1}{f_P}$: $\frac{1}{v_P} - \frac{1}{-15} = \frac{1}{10} \Rightarrow \frac{1}{v_P} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$. So, $v_P = +30$ cm. The image formed by lens P is real, 30 cm to the right of P. Now, for lens Q (convex, $f_Q = +

Q: What are the options for this JEE Main Physics question?

A. virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB · B. real, formed at 7.5 cm right of lens Q, with a size same than that of AB · C. formed at infinity · D. real, formed at 7 cm right of lens Q, with a size smaller than that of AB

Question

Q.41. An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is

FirstInTest · Accepted Answer

For lens P (convex, $f_P = +10$ cm):
Object distance $u_P = -15$ cm (left of P).
Using lens formula $\frac{1}{v_P} - \frac{1}{u_P} = \frac{1}{f_P}$:
$\frac{1}{v_P} - \frac{1}{-15} = \frac{1}{10} \Rightarrow \frac{1}{v_P} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$. So, $v_P = +30$ cm. The image formed by lens P is real, 30 cm to the right of P.

Now, for lens Q (convex, $f_Q = +15$ cm):
Lens Q is placed 15 cm right of lens P. The image from P acts as the object for Q. The distance of this image from P is 30 cm to its right. Since Q is 15 cm right of P, the object distance for Q, $u_Q = +(30 - 15) = +15$ cm. This means the object for Q is virtual, located 15 cm to the right of Q.
Using lens formula for Q:
$\frac{1}{v_Q} - \frac{1}{u_Q} = \frac{1}{f_Q} \Rightarrow \frac{1}{v_Q} - \frac{1}{+15} = \frac{1}{15} \Rightarrow \frac{1}{v_Q} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15}$. So, $v_Q = +\frac{15}{2} = +7.5$ cm.
The final image is real, 7.5 cm to the right of lens Q.

Magnification for P: $m_P = \frac{v_P}{u_P} = \frac{+30}{-15} = -2$.
Magnification for Q: $m_Q = \frac{v_Q}{u_Q} = \frac{+7.5}{+15} = +0.5$.
Total magnification $m = m_P m_Q = (-2)(+0.5) = -1$. The final image is inverted and has the same size as the object.

Therefore, the final image is real, formed at 7.5 cm right of lens Q, with a size same as that of AB.

Answer

virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB

Answer

real, formed at 7.5 cm right of lens Q, with a size same than that of AB

Answer

formed at infinity

Answer

real, formed at 7 cm right of lens Q, with a size smaller than that of AB

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