uj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/ee58bbf1a9a4.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 4th April Evening Shift Physics - Geometrical Optics Question 39 English Explanation">

$$\begin{aligned} &\text { For Convex mirror }\\ &\begin{aligned} & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{30}=\frac{1}{30} \\ & \frac{1}{\mathrm{v}}=\frac{2}{30}=\frac{1}{15} \Rightarrow \mathrm{v}=15 \mathrm{~cm} \end{aligned} \end{aligned}$$

Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coinside,

Therefore distance between both mirrors

$$=30-22.5=7.5 \mathrm{~cm}$$

Correct Answer : Option 2

19:T5bf,

JEE Main 2025 (Online) 4th April Evening Shift Physics - Geometrical Optics Question 39 Hindi Explanation

$$\begin{aligned} &\text { उत्तल दर्पण के लिए }\\\begin{aligned} & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{30}=\frac{1}{30} \\ & \frac{1}{\mathrm{v}}=\frac{2}{30}=\frac{1}{15} \Rightarrow \mathrm{v}=15 \mathrm{~cm} \end{aligned} \end{aligned}$$

उत्तल दर्पण द्वारा बना प्रतिबिम्ब वस्तु से 45 cm की दूरी पर है इसलिए समतल दर्पण को वस्तु से 22.5 cm की दूरी पर मध

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